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Revision History for A192455 (Underlined text is an addition; strikethrough text is a deletion.)

Showing entries 1-10 | older changes
A192455 G.f. A(x) satisfies: 1 = Sum_{n>=0} x^n*A(-x)^A001650(n+1), where A001650 is defined by "n appears n times (n odd).".
(history; published version)
#12 by Russ Cox at Fri Mar 30 18:37:27 EDT 2012
AUTHOR

_Paul D. Hanna (pauldhanna(AT)juno.com), _, Jul 14 2011

Discussion
Fri Mar 30 18:37
OEIS Server: https://oeis.org/edit/global/213
#11 by Joerg Arndt at Fri Jul 15 03:28:34 EDT 2011
STATUS

proposed

approved

#10 by Paul D. Hanna at Thu Jul 14 23:39:17 EDT 2011
STATUS

editing

proposed

#9 by Paul D. Hanna at Thu Jul 14 23:39:07 EDT 2011
NAME

G.f. A(x) satisfies: 1 = Sum_{n>=0} x^n*A(-x)^A001650(n+1), where A001650 is defined by "n appears n times (n odd).".

COMMENTS

Compare the g.f. to a g.f. C(x) of the Catalan numbers: 1 = Sum_{n>=0} x^n*C(-x)^(2*n+1).

FORMULA

G.f. satisfies: 1-x = Sum_{n>=1} x^(n^2)* () * (1-x^(2*n-1))* )) * A(-x)^(2*n-1).

EXAMPLE

G.f.: A(x) = 1 + x + 2*x^2 + 7*x^3 + 27*x^4 + 112*x^5 + 492*x^6 +...

PROG

((PARI) {a(n)=local(A=[1]); for(i=1, n, A=concat(A, 0); A[#A]=polcoeff(sum(m=1, #A, (-x)^m*Ser(A)^(1+2*sqrtint(m-1)) ), #A)); if(n<0, 0, A[n+1])}

CROSSREFS

Cf. A193039, A193040, A193050, A001650.

STATUS

proposed

editing

#8 by Paul D. Hanna at Thu Jul 14 21:18:15 EDT 2011
STATUS

editing

proposed

#7 by Paul D. Hanna at Thu Jul 14 21:17:29 EDT 2011
NAME

allocated for Paul D. Hanna

G.f. A(x) satisfies: 1 = Sum_{n>=0} x^n*A(-x)^A001650(n+1), where A001650 is defined by "n appears n times (n odd).".

DATA

1, 1, 2, 7, 27, 112, 492, 2249, 10580, 50885, 249067, 1236602, 6212563, 31523293, 161317863, 831615320, 4314659345, 22512421092, 118052038100, 621825506334, 3288597601727, 17455485596492, 92958082866815, 496535775228131, 2659574264906443

OFFSET

0,3

COMMENTS

Compare the g.f. to a g.f. C(x) of the Catalan numbers: 1 = Sum_{n>=0} x^n*C(-x)^(2*n+1).

FORMULA

G.f. satisfies: 1-x = Sum_{n>=1} x^(n^2)* (1-x^(2*n-1))* A(-x)^(2*n-1).

EXAMPLE

G.f.: A(x) = 1 + x + 2*x^2 + 7*x^3 + 27*x^4 + 112*x^5 + 492*x^6 +...

The g.f. satisfies:

1 = A(-x) + x*A(-x)^3 + x^2*A(-x)^3 + x^3*A(-x)^3 + x^4*A(-x)^5 + x^5*A(-x)^5 + x^6*A(-x)^5 + x^7*A(-x)^5 + x^8*A(-x)^5 + x^9*A(-x)^7 +...+ x^n*A(-x)^A001650(n+1) +...

where A001650 begins: [1, 3,3,3, 5,5,5,5,5, 7,7,7,7,7,7,7, 9,...].

The g.f. also satisfies:

1-x = (1-x)*A(-x) + x*(1-x^3)*A(-x)^3 + x^4*(1-x^5)*A(-x)^5 + x^9*(1-x^7)*A(-x)^7 + x^16*(1-x^9)*A(-x)^9 +...

PROG

(PARI) {a(n)=local(A=[1]); for(i=1, n, A=concat(A, 0); A[#A]=polcoeff(sum(m=1, #A, (-x)^m*Ser(A)^(1+2*sqrtint(m-1)) ), #A)); if(n<0, 0, A[n+1])}

CROSSREFS

Cf. A193039, A193040, A001650.

KEYWORD

allocated

nonn

AUTHOR

Paul D. Hanna (pauldhanna(AT)juno.com), Jul 14 2011

STATUS

approved

editing

#6 by Paul D. Hanna at Thu Jul 14 21:11:16 EDT 2011
NAME

allocated for Paul D. Hanna

KEYWORD

recycled

allocated

#5 by T. D. Noe at Thu Jul 14 19:56:15 EDT 2011
STATUS

editing

approved

#4 by T. D. Noe at Thu Jul 14 19:56:11 EDT 2011
NAME

a(n) is the smallest k such that prime(n) divides prime(k)+ prime(k+1).

DATA

2, 3, 1, 8, 25, 9, 11, 21, 19, 69, 24, 29, 46, 23, 60, 115, 51, 111, 32, 82, 129, 185, 132, 71, 106, 155, 63, 116, 84, 203, 54, 77, 58, 145, 108, 87, 289, 93, 67, 443, 254, 460, 292, 76, 350, 300, 447, 86, 397, 124, 284, 808, 128, 335, 136, 547, 742, 361, 571, 475

OFFSET

1,1

COMMENTS

See the sequence A188815 for prime(k).

EXAMPLE

a(4) = 8 because prime(4) = 7 divides prime(8) + prime(9) = 19 + 23 = 42.

MAPLE

with(numtheory):for n from 1 to 60 do:p1:=ithprime(n):id:=0:for k from 1 to 1000 while(id=0) do: x:=ithprime(k)+ ithprime(k+1):if irem(x, p1)=0 then id:=1:printf(`%d, `, k):else fi:od:od:

CROSSREFS

Cf.A188815.

KEYWORD

nonn,changed

recycled

AUTHOR

Michel Lagneau (mn.lagneau2(AT)orange.fr), Jul 01 2011

STATUS

proposed

editing

#3 by Michel Lagneau at Fri Jul 01 16:05:01 EDT 2011
COMMENTS

See the sequence A188815 for prime(k).

EXAMPLE

a(4) = 8 because prime(4) = 7 divides prime(8) + prime(9) = 19 + 23 = 42.

CROSSREFS

Cf.A188815.

Discussion
Wed Jul 13 14:05
T. D. Noe: I do not see why we need this sequence if we already have A188815.

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