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a(n) = (3*9^n - 5^n)/2.
+10
3
1, 11, 109, 1031, 9529, 87011, 789349, 7135391, 64374769, 580154171, 5225293789, 47047175351, 423522234409, 3812188390931, 34312136924629, 308821439352911, 2779453989332449, 25015391079773291, 225140045596865869, 2026268039766324071, 18236450504869572889, 164128245278689437251
OFFSET
0,2
COMMENTS
Binomial transform of A165148. Inverse binomial transform of A165150.
FORMULA
a(n) = 14*a(n-1) - 45*a(n-2) for n > 1; a(0) = 1, a(1) = 11.
G.f.: (1 - 3*x)/((1 - 5*x)*(1 - 9*x)).
E.g.f.: exp(5*x)*(3*exp(4*x) - 1)/2. - Stefano Spezia, Mar 21 2023
MAPLE
A165149:=n->(3*9^n-5^n)/2; seq(A165149(n), n=0..30); # Wesley Ivan Hurt, Dec 10 2013
MATHEMATICA
Table[(3*9^n - 5^n)/2, {n, 0, 30}] (* Wesley Ivan Hurt, Dec 10 2013 *)
PROG
(Magma) [ (3*9^n-5^n)/2: n in [0..18] ];
CROSSREFS
KEYWORD
nonn,easy
AUTHOR
Klaus Brockhaus, Sep 15 2009
STATUS
approved
a(n) = (3*7^n-3^n)/2.
+10
2
1, 9, 69, 501, 3561, 25089, 176109, 1234221, 8643921, 60520569, 423683349, 2965901541, 20761665081, 145332718449, 1017332217789, 7121335090461, 49849374331041, 348945706410729, 2442620203155429, 17098342196928981
OFFSET
0,2
COMMENTS
Partial sums are in A016138.
Binomial transform of A016129. Inverse binomial transform of A165148.
FORMULA
a(n) = 10*a(n-1)-21*a(n-2) for n > 1; a(0) = 1, a(1) = 9.
G.f.: (1-x)/((1-3*x)*(1-7*x)).
MATHEMATICA
LinearRecurrence[{10, -21}, {1, 9}, 25] (* Paolo Xausa, Apr 22 2024 *)
PROG
(Magma) [ (3*7^n-3^n)/2: n in [0..19] ];
CROSSREFS
KEYWORD
nonn,easy
AUTHOR
Klaus Brockhaus, Sep 15 2009
STATUS
approved
Triangle read by rows: T(n,k) is the number of 2 X 2 matrices over Z(n) having determinant congruent to k mod n, 1 <= n, 0 <= k <= n-1.
+10
2
1, 10, 6, 33, 24, 24, 88, 48, 72, 48, 145, 120, 120, 120, 120, 330, 144, 240, 198, 240, 144, 385, 336, 336, 336, 336, 336, 336, 736, 384, 576, 384, 672, 384, 576, 384, 945, 648, 648, 864, 648, 648, 864, 648, 648, 1450, 720, 1200, 720, 1200, 870, 1200, 720, 1200, 720
OFFSET
1,2
COMMENTS
The n-th row is {T(n,0),T(n,1),...,T(n,n-1)}.
Let m denote the prime power p^e.
T(m,0) = A020478(m) = (p^(e+1) + p^e-1)*p^(2*e-1).
T(m,1) = A000056(m) = (p^2-1)*p^(3*e-2).
T(prime(n),1) = A127917(n).
Sum_{k=1..n-1} T(n,k) = A005353(n).
T(n,1) = n*A007434(n) for n>=1 because A000056(n) = n*Jordan_Function_J_2(n).
T(2^n,1) = A083233(n) = A164640(2n) for n>=1. Proof: a(n):=T(2^n,1); a(1)=6, a(n)=8*a(n-1); A083233(1)=6 and A083233(n) is a geometric series with ratio 8 (because of its g.f.), too; A164640 = {b(1)=1, b(2)=6, b(n)=8*b(n-2)}.
T(2^n,0) = A165148(n) for n>=0, because 2*T(2^n,0) = (3*2^n-1)*4^n.
T(2^e,2) = A003951(e) for 2 <= e. Proof: T(2^e,2) = 9*8^(e-1) is a series with ratio 8 and initial term 72, as A003951(2...inf) is.
Working with consecutive powers of a prime p, we need a definition (0 <= i < e):
N(p^e,i):=#{k: 0 < k < p^e, gcd(k,p^e) = p^i} = (p-1)*p^(e-1-i). We say that these k's belong to i (respect to p^e). Note that N(p^e,0) = EulerPhi(p^e), and if 0 < k < p^e then gcd(k,p^e) = gcd(k,p^(e+1)). Let T(p^e,[i]) denote the common value of T(p^e,k)'s, where k's belong to i (q.v.PROGRAM); for example, T(p^e,[0]) = T(p^e,1). The number of the 2 X 2 matrices over Z(p^e), T(p^e,0) + Sum_{i=0..e-1} T(p^e,[i])*N(p^e,i) = p^(4e) will be useful.
On the hexagon property: Let prime p be given and let T(p^e,[0]), T(p^e,[1]), T(p^e,[2]), ..., T(p^e,[e-2]), T(p^e,[e-1]) form the e-th row of a Pascal-like triangle, e>=1. Let denote X(r,s) an element of the triangle and its value T(p^r,[s]). Let positive integers a and b given, so that the entries A(m-a,n-b), B(m-a,n), C(m,n+a), D(m+b,n+a), E(m+b,n), F(m,n-b) of the triangle form a hexagon spaced around T(p^m,[n]); if a=b=1 then they surround it. If A*C*E = B*D*F, then we say that the triangle T(.,.) has the "hexagon property". (In the case of binomial coefficients X(r,s) = COMB(r,s), the "hexagon property" holds (see [Gupta]) and moreover gcd(A,C,E) = gcd(B,D,F) (see [Hitotumatu & Sato]).)
Corollary 2.2 in [Brent & McKay] says that, for the d X d matrices over Z(p^e), (mutatis mutandis) T_d(p^e,0) = K*(1-P(d+e-1)/P(e-1)) and T_d(p^e,[i]) = K*(q^e)*((1-q^d)/(1-q))*P(d+i-1)/P(i), where q=1/p, K=(p^e)^(d^2), P(t) = Product_{j=1..t} (1-q^j), P(0):=1. (For the case d=2, we have T(p^e,[i]) = (p+1)*(p^(i+1)-1)*p^(3*e-i-2).) Due to [Brent & McKay], it can be simply proved that for d X d matrices the "hexagon property" is true. The formulation implies an obvious generalization: For the entries A(r,u), B(r,v), C(s,w), D(t,w), E(t,v), F(s,u) of the T_d(.,.)-triangle, a hexagon-like property A*C*E = B*D*F holds. This is false in general for the COMB(.,.)-triangle.
Another (rotated-hexagon-like) property: for the entries A(m-b1,n), B(m-a1,n+c2), C(m+a2,n+c2), D(m+b2,n), E(m+a2,n-c1), F(m-a1,n-c1) of the T_d(.,.)-triangle, the property A*C*E = B*D*F holds, if and only if 2*(a1 + a2) = b1 + b2. This is also in general false for COMB(.,.)-triangle.
LINKS
Richard P. Brent and Brendan D. McKay, Determinants and ranks of random matrices over Z_m, Discrete Mathematics 66 (1987) pp. 35-49.
A. K. Gupta, Generalized hidden hexagon squares, The Fibonacci Quarterly, Vol 12, Number 1, Feb.1974, pp. 45-46.
S. Hitotumatu, D. Sato, Star of David theorem (I), The Fibonacci Quarterly, Vol 13, Number 1, Feb.1975, p. 70.
FORMULA
T(a*b,k) = T(a,(k mod a))*T(b,(k mod b)) if gcd(a,b) = 1.
Sum_{k=1..n-1, gcd(k,n)=1} T(n,k) = A000252(n). - Andrew Howroyd, Jul 16 2018
EXAMPLE
From Andrew Howroyd, Jul 16 2018: (Start)
Triangle begins:
1;
10, 6;
33, 24, 24;
88, 48, 72, 48;
145, 120, 120, 120, 120;
330, 144, 240, 198, 240, 144;
385, 336, 336, 336, 336, 336, 336;
736, 384, 576, 384, 672, 384, 576, 384;
945, 648, 648, 864, 648, 648, 864, 648, 648;
... (End)
PROG
(Other) . (* computing T(p^e, k) ; p=prime, 1<=e, 0<=k<p^e, elementary approach *)
. (1) F := (p-1)*p^(e-1)
. (2) u := [ u(0), u(1), ..., u(p^e-1) ] vector, where
. (21) u(0) := p^e + e*F, and
. (22) FOR x := 1 TO p^e-1
. (22) u(x) := (i+1)*F, where GCD(x, p^e)= p^i
. (22) ENDFOR
. (3) T(p^e, k):= ScalarProduct( u, kTimesCyclicRightShift(u) )
(PARI)
S(p, e)={my(u=vector(p^e)); my(t=(p-1)*p^(e-1)); u[1] = p^e + e*t; for(j=1, p^e-1, u[j+1] = t*(1+valuation(j, p))); vector(#u, k, sum(j=0, #u-1, u[j + 1]*u[(j+k-1) % #u + 1]))}
T(n)={my(f=factor(n), v=vector(n, i, 1)); for(i=1, #f~, my(r=S(f[i, 1], f[i, 2])); for(j=0, #v-1, v[j + 1] *= r[j % #r + 1])); v}
for(n=1, 10, print(T(n))); \\ Andrew Howroyd, Jul 16 2018
CROSSREFS
Column k=0 is A020478.
Column k=1 is A000056.
Row sums are A005353.
KEYWORD
mult,nonn,tabl
AUTHOR
Erdos Pal, Oct 03 2010
EXTENSIONS
Terms a(24)-a(55) from b-file by Andrew Howroyd, Jul 16 2018
STATUS
approved

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