cp's OEIS Frontend

This is a front-end for the Online Encyclopedia of Integer Sequences, made by Christian Perfect. The idea is to provide OEIS entries in non-ancient HTML, and then to think about how they're presented visually. The source code is on GitHub.

Showing 1-3 of 3 results.

A290581 Column 1 of triangle A290580.

Original entry on oeis.org

1, 20, 364, 7028, 148752, 3471192, 89097664, 2503362488, 76575071488, 2536513162508, 90532686154752, 3465845396598540, 141726054915248128, 6167370619705004144, 284635248765764878336, 13889365886508877963184, 714612366728939248091136, 38667882915659893417754820, 2195406208657284278474506240, 130509259754593318496376665060, 8107405596523654695095077175296, 525358015896796804145274761270600
Offset: 1

Views

Author

Paul D. Hanna, Aug 07 2017

Keywords

Comments

a(n) = A290580(n+2,1) for n>=1.

Crossrefs

Programs

  • PARI
    /* As column 1 of triangle A290580 */
    { A290580(n,k) = my(W=1,E=1, S=x,C=1,D=1); for(i=0,n,
    S = intformal(C*D +x*O(x^n)) ;
    C = 1 - intformal(S*D) ; D = 1 - m*intformal(S*C) ;
    E = subst( (1 + S)/C,m,1-m) ) ;
    for(i=0,n, W = subst(E,x,x*W));
    n!*polcoeff(polcoeff(W, n,x), k,m) }
    for(n=1,25, print1( A290580(n+2,1), ", "))

Formula

a(n) ~ c * n^(n+2), where c = (exp(5) - 4*exp(3) - exp(1))/16 = 4.0845455988354304513775733719295... - Vaclav Kotesovec, Aug 21 2017

A291214 Column 2 of triangle A290580.

Original entry on oeis.org

1, 112, 5868, 250128, 10020912, 399379728, 16255733440, 684615750832, 30031767680256, 1376568893633760, 66017645596167168, 3313241694194681184, 173934275433107845120, 9543378596912872361440, 546711252967087466397696, 32663132242303127521217184
Offset: 1

Views

Author

Vaclav Kotesovec, Aug 21 2017

Keywords

Crossrefs

Programs

  • PARI
    /* As column 2 of triangle A290580 */
    { A290580(n, k) = my(W=1, E=1, S=x, C=1, D=1); for(i=0, n,
    S = intformal(C*D +x*O(x^n)) ;
    C = 1 - intformal(S*D) ; D = 1 - m*intformal(S*C) ;
    E = subst( (1 + S)/C, m, 1-m) ) ;
    for(i=0, n, W = subst(E, x, x*W));
    n!*polcoeff(polcoeff(W, n, x), k, m) }
    for(n=1, 25, print1( A290580(n+4, 2), ", ")) \\ after Paul D. Hanna

Formula

a(n) ~ c * n^(n+5), where c = (exp(1) + 8*exp(3) + 14*exp(5) - 8*exp(7) + exp(9))/512 = 3.06876067343310165753640903985063833178947434...

A290579 E.g.f. satisfies: A(x) = (1 + sin(x*A(x))) / cos(x*A(x)).

Original entry on oeis.org

1, 1, 3, 17, 145, 1661, 23947, 416765, 8504865, 199200985, 5268047411, 155282897065, 5048546868209, 179483438440021, 6927036355244475, 288439692143197397, 12889578930617144641, 615304922810783744561, 31249797537098034966499, 1682478707717022082757441, 95721167135792642044679505, 5738213812249917154871122861, 361520344958259229318337826923, 23881324878748364254507881173677
Offset: 0

Views

Author

Paul D. Hanna, Aug 06 2017

Keywords

Comments

Equals row sums of irregular triangle A290580.
A diagonal in square array A185414.

Examples

			E.g.f.: A(x) = 1 + x + 3*x^2/2! + 17*x^3/3! + 145*x^4/4! + 1661*x^5/5! + 23947*x^6/6! + 416765*x^7/7! + 8504865*x^8/8! +...
such that A(x) = (1 + sin(x*A(x))) / cos(x*A(x)).
RELATED SERIES.
(1 + sin(x)) / cos(x) = 1 + x + x^2/2! + 2*x^3/3! + 5*x^4/4! + 16*x^5/5! + 61*x^6/6! + 272*x^7/7! + 1385*x^8/8! +...+ A000111(n)*x^n/n! +...
where A000111 is the sequence of zigzag numbers.
Given e.g.f. A(x), then the logarithm of A(x) begins:
log(A(x)) = x + 2*x^2/2! + 10*x^3/3! + 80*x^4/4! + 880*x^5/5! + 12336*x^6/6! + 210320*x^7/7! + 4225024*x^8/8! +...+ A185414(n,n)*x^n/n! +...
		

Crossrefs

Programs

  • PARI
    {a(n) = my(A=1); for(i=0,n, A = (1 + sin(x*A)) / cos(x*A +x*O(x^n)) ); n!*polcoeff(A,n)}
    for(n=0,25, print1(a(n),", "))
    
  • PARI
    {a(n) = my(X=x +O(x^(n+2))); A = (1/x) * serreverse( x*cos(X)/(1 + sin(X)) ); n!*polcoeff(A,n)}
    for(n=0,25, print1(a(n),", "))

Formula

E.g.f.: A(x) = (1/x) * Series_Reversion( x*cos(x)/(1 + sin(x)) ).
E.g.f. satisfies: A(x) = Sum_{n>=0} A000111(n) * x^n * A(x)^n / n!.
E.g.f.: A(x) = exp( sum_{n>=1} A185414(n,n)*x^n/n! ).
a(n) = A185414(n,n+1) for n>=1.
a(n) = [x^n/n!] ( (1 + sin(x)) / cos(x) )^(n+1) / (n+1) for n>=0.
a(n) ~ n^(n-1) / (exp(n) * r^(n+1/2)), where r = 0.3263879708167851846572540343... is the root of the equation r = 1 - sin(sqrt(r*(2-r))). - Vaclav Kotesovec, Aug 20 2017
Showing 1-3 of 3 results.