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Search: a243941 -id:a243941
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a(n) = (6*n)^2.
+10
7
0, 36, 144, 324, 576, 900, 1296, 1764, 2304, 2916, 3600, 4356, 5184, 6084, 7056, 8100, 9216, 10404, 11664, 12996, 14400, 15876, 17424, 19044, 20736, 22500, 24336, 26244, 28224, 30276, 32400, 34596, 36864, 39204, 41616, 44100, 46656, 49284, 51984, 54756, 57600, 60516, 63504, 66564, 69696, 72900
OFFSET
0,2
COMMENTS
Areas A of two classes of triangles with integer sides (a,b,c) where a = 9k, b=10k and c = 17k, or a = 3k, b = 25k and c = 26k for k=0,1,2,... These areas are given by Heron's formula A = sqrt(s(s-a)(s-b)(s-c)) = (6k)^2, with the semiperimeter s = (a+b+c)/2. This sequence is a subsequence of A188158. - Michel Lagneau, Oct 11 2013
Sequence found by reading the line from 0, in the direction 0, 36, ..., in the square spiral whose vertices are the generalized 20-gonal numbers A218864. - Omar E. Pol, May 13 2018.
FORMULA
From Ilya Gutkovskiy, Jun 09 2016: (Start)
O.g.f.: 36*x*(1 + x)/(1 - x)^3.
E.g.f.: 36*x*(1 + x)*exp(x).
a(n) = 3*a(n-1) - 3*a(n-2) + a(n-3).
Sum_{n>=1} 1/a(n) = Pi^2/216 = A086726. (End)
Product_{n>=1} a(n)/A136017(n) = Pi/3. - Fred Daniel Kline, Jun 09 2016
a(n) = t(9*n) - 9*t(n), where t(i) = i*(i+k)/2 for any k. Special case (k=1): a(n) = A000217(9*n) - 9*A000217(n). - Bruno Berselli, Aug 31 2017
a(n) = 36*A000290(n) = 18*A001105(n) = 12*A033428 = 9*A016742(n) = 6*A033581(n) = 4*A016766(n) = 3*A135453(n) = 2*A195321(n). - Omar E. Pol, Jun 07 2018
Sum_{n>=1} (-1)^(n+1)/a(n) = Pi^2/432. - Amiram Eldar, Jun 27 2020
From Amiram Eldar, Jan 25 2021: (Start)
Product_{n>=1} (1 + 1/a(n)) = sinh(Pi/6)/(Pi/6).
Product_{n>=1} (1 - 1/a(n)) = sin(Pi/6)/(Pi/6) = 3/Pi (A089491). (End)
MATHEMATICA
(6*Range[0, 40])^2 (* or *) LinearRecurrence[{3, -3, 1}, {0, 36, 144}, 40] (* Harvey P. Dale, Dec 25 2017 *)
Table[36 n^2, {n, 0, 45}] (* Omar E. Pol, Jun 07 2018 *)
PROG
(Magma) [(6*n)^2: n in [0..40]]; // Vincenzo Librandi, May 03 2011
(PARI) a(n)=36*n^2 \\ Charles R Greathouse IV, Jun 10 2016
CROSSREFS
Cf. similar sequences of the type k*n^2: A000290 (k=1), A001105 (k=2), A033428 (k=3), A016742 (k=4), A033429 (k=5), A033581 (k=6), A033582 (k=7), A139098 (k=8), A016766 (k=9), A033583 (k=10), A033584 (k=11), A135453 (k=12), A152742 (k=13), A144555 (k=14), A064761 (k=15), A016802 (k=16), A244630 (k=17), A195321 (k=18), A244631 (k=19), A195322 (k=20), A064762 (k=21), A195323 (k=22), A244632 (k=23), A195824 (k=24), A016850 (k=25), A244633 (k=26), A244634 (k=27), A064763 (k=28), A244635 (k=29), A244636 (k=30).
KEYWORD
nonn,easy
STATUS
approved
Number of decompositions of 12*n into the sum of two (not necessarily distinct) twin prime pairs.
+10
0
0, 1, 1, 2, 1, 2, 1, 2, 1, 2, 1, 2, 2, 2, 2, 0, 2, 1, 3, 3, 1, 2, 1, 3, 2, 2, 2, 3, 1, 3, 1, 2, 3, 3, 6, 2, 3, 1, 2, 4, 3, 4, 4, 1, 3, 2, 3, 5, 2, 7, 1, 3, 2, 2, 5, 2, 5, 2, 3, 2, 2, 3, 5, 3, 4, 1, 0, 3, 1, 6, 2, 3, 3, 1, 5, 2, 5, 3, 3, 4, 1, 4
OFFSET
1,4
COMMENTS
In the 1980's, Liang conjectured that (6n)^2 = p_1 + p_2 + p_3 + p_4, where (p_1, p_2) and (p_3, p_4) are twin prime pairs. See reference for more details.
It seems there are at least 2 solutions for the decompositions when n > 701.
If the two twin prime pairs are required to be distinct, the sequence is A187759.
REFERENCES
Liang Ding Xiang, Problem 93#, Bulletin of Mathematics (Wuhan), 6(1992),41. ISSN 0488-7395.
EXAMPLE
a(736) = 2 because 12*736 = 197 + 199 + 4217 + 4219 = 857 + 859 + 3557 + 3559, so there are 2 ways of expressing 12*n as the sum of two twin prime pairs.
PROG
(PARI) v=select(p->isprime(p-2)&&p>5, primes(200))\6; l=List(); for(i=1, #v, if(2*v[i]<100, listput(l, 2*v[i])); for(j=i+1, #v, if((v[i]+v[j])<100, listput(l, v[i]+v[j])))); l1=vecsort(l); k=1; for(i=1, 100, s=sum(j=k, #l1, l1[j]==i); print1(s", "); k+=s) \\ Lear Young, Jun 16 2014
(PARI) v=select(p->isprime(p-2)&&p>5, primes(110))\6; for(i=1, 99, print1(sum(j=1, #v, vecsearch(v, i-v[j])>0 && i-v[j]>=v[j])", ")) \\ change i-v[j]>=v[j] to i-v[j]>v[j] is A187759. Lear Young, Jun 16 2014
CROSSREFS
KEYWORD
nonn
AUTHOR
Lear Young, Jun 16 2014
STATUS
approved

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