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 1-8 of 8 results.

A232606 G.f. A(x) satisfies: the sum of the coefficients of x^k, k=0..n, in A(x)^n equals (2*n)!^2/n!^4, the square of the central binomial coefficients (A000984), for n>=0.

Original entry on oeis.org

1, 3, 10, 42, 221, 1379, 9678, 73666, 594326, 5007958, 43641702, 390632678, 3573598539, 33289289533, 314871186248, 3017358158132, 29242725947318, 286209134234602, 2825613061237808, 28111283170770480, 281598654896870051, 2838309465080014489, 28767973963085929656, 293059625830028920012
Offset: 0

Views

Author

Paul D. Hanna, Nov 26 2013

Keywords

Comments

Compare to: Sum_{k=0..n} [x^k] 1/(1-x)^n = (2*n)!/n!^2 = A000984(n).
a(n+1)/a(n) tends to 11.3035... - Vaclav Kotesovec, Jan 23 2014

Examples

			G.f.: A(x) = 1 + 3*x + 10*x^2 + 42*x^3 + 221*x^4 + 1379*x^5 + 9678*x^6 +...
ILLUSTRATION OF INITIAL TERMS.
If we form an array of coefficients of x^k in A(x)^n, n>=0, like so:
A^0: [1], 0,   0,    0,     0,      0,       0,        0,         0, ...;
A^1: [1,  3], 10,   42,   221,   1379,    9678,    73666,    594326, ...;
A^2: [1,  6,  29], 144,   794,   4924,   33814,   251544,   1988885, ...;
A^3: [1,  9,  57,  333], 1989,  12669,   86935,   639123,   4979499, ...;
A^4: [1, 12,  94,  636,  4157], 27728,  193504,  1423120,  11006058, ...;
A^5: [1, 15, 140, 1080,  7730,  54538], 391970,  2915490,  22558825, ...;
A^6: [1, 18, 195, 1692, 13221,  99102,  739547], 5612016,  43767477, ...;
A^7: [1, 21, 259, 2499, 21224, 169232, 1317722, 10267666], 81223912, ...;
A^8: [1, 24, 332, 3528, 32414, 274792, 2238492, 17990904, 145096413], ...; ...
then the sum of the coefficients of x^k, k=0..n, in A(x)^n (shown above in brackets) equals the square of the central binomial coefficients:
1^1 = 1;
2^2 = 1 + 3;
6^2 = 1 +  6 +  29;
20^2 = 1 +  9 +  57 +  333;
70^2 = 1 + 12 +  94 +  636 +  4157;
252^2 = 1 + 15 + 140 + 1080 +  7730 +  54538;
924^2 = 1 + 18 + 195 + 1692 + 13221 +  99102 +  739547;
3432^2 = 1 + 21 + 259 + 2499 + 21224 + 169232 + 1317722 + 10267666; ...
RELATED SERIES.
From a main diagonal in the above array we can derive sequence A232607:
[1/1, 6/2, 57/3, 636/4, 7730/5, 99102/6, 1317722/7, 17990904/8, ...] =
[1, 3, 19, 159, 1546, 16517, 188246, 2248863, 27844369, 354576634, ...];
from which we can form the series G(x) = A(x*G(x)):
G(x) = 1 + 3*x + 19*x^2 + 159*x^3 + 1546*x^4 + 16517*x^5 + 188246*x^6 +...
such that
(G(x) + x*G'(x)) / (G(x) - x*G(x)^2) = 1 + 2^2*x + 6^2*x^2 + 20^2*x^3 + 70^2*x^4 + 252^2*x^5 +...+ A000984(n)^2*x^n +...

Links

Crossrefs

Cf. A232683, A232607, A232687, A000984, A002894.

Programs

  • Mathematica
    terms = 24; a[0] = 1; A[x_] = Sum[a[n]*x^n, {n, 0, terms - 1}];
c[n_] := Sum[Coefficient[B[x], x, k], {k, 0, n}] == (2*n)!^2/n!^4 // Solve // First;
Do[B[x_] = A[x]^n + O[x]^(n+1) // Normal; A[x_] = (A[x] /. c[n]) + O[x]^terms, {n, 0, terms-1}];
CoefficientList[A[x], x] (* Jean-François Alcover, Jan 14 2018 *)
  • PARI
    /* By Definition: */
{a(n)=if(n==0, 1, ((2*n)!^2/n!^4 - sum(k=0, n, polcoeff(sum(j=0, min(k, n-1), a(j)*x^j)^n + x*O(x^k), k)))/n)}
for(n=0,20,print1(a(n)*1!,", "))
  • PARI
    /* Faster, using series reversion: */
{a(n)=local(CB2=sum(k=0,n,binomial(2*k,k)^2*x^k)+x*O(x^n), G=1+x*O(x^n));
for(i=1,n,G = 1 + intformal( (CB2-1)*G/x - CB2*G^2));polcoeff(x/serreverse(x*G),n)}
for(n=0,30,print1(a(n),", "))

Formula

Given g.f. A(x), Sum_{k=0..n} [x^k] A(x)^n = (2*n)!^2/n!^4 = A000984(n)^2.
Given g.f. A(x), let G(x) = A(x*G(x)) then (G(x) + x*G'(x)) / (G(x) - x*G(x)^2) = Sum_{n>=0} (2*n)!^2/n!^4 * x^n.

A232687 G.f. A(x) satisfies: the sum of the coefficients of x^k, k=0..n, in A(x)^n equals Sum_{k=0..n} C(n,k)^3 = A000172(n) (Franel numbers), for n>=0.

Original entry on oeis.org

1, 1, 3, 7, 20, 66, 244, 980, 4182, 18674, 86353, 410541, 1996214, 9888844, 49760925, 253767097, 1309154825, 6822023553, 35865392690, 190038440422, 1014015337209, 5444707218851, 29401289997403, 159584901816255, 870267544114291, 4766246752344215, 26206635040151511
Offset: 0

Views

Author

Paul D. Hanna, Dec 05 2013

Keywords

Comments

Compare to: Sum_{k=0..n} [x^k] 1/(1-x)^n = Sum_{k=0..n} C(n,k)^2 = (2*n)!/n!^2.
a(n+1)/a(n) tends to 6.0295... - Vaclav Kotesovec, Jan 22 2014

Examples

			G.f.: A(x) = 1 + x + 3*x^2 + 7*x^3 + 20*x^4 + 66*x^5 + 244*x^6 + 980*x^7 +...
ILLUSTRATION OF INITIAL TERMS.
If we form an array of coefficients of x^k in A(x)^n, n>=0, like so:
A^0: [1],0,  0,   0,    0,    0,     0,      0,      0, ...;
A^1: [1, 1], 3,   7,   20,   66,   244,    980,   4182, ...;
A^2: [1, 2,  7], 20,   63,  214,   789,   3124,  13112, ...;
A^3: [1, 3, 12,  40], 138,  492,  1848,   7326,  30531, ...;
A^4: [1, 4, 18,  68,  255], 960,  3716,  14920,  62295, ...;
A^5: [1, 5, 25, 105,  425, 1691], 6785,  27805, 117165, ...;
A^6: [1, 6, 33, 152,  660, 2772, 11560], 48588, 207774, ...;
A^7: [1, 7, 42, 210,  973, 4305, 18676,  80746],351792, ...;
A^8: [1, 8, 52, 280, 1378, 6408, 28916, 128808, 573311], ...; ...
then the sum of the coefficients of x^k, k=0..n, in A(x)^n (shown above in brackets) equals Sum_{k=0..n} C(n,k)^3 = A000172(n):
A000172(0) = 1 = 1;
A000172(1) = 1 + 1 = 2;
A000172(2) = 1 + 2 +  7 = 10;
A000172(3) = 1 + 3 + 12 +  40 = 56;
A000172(4) = 1 + 4 + 18 +  68 + 255 = 346;
A000172(5) = 1 + 5 + 25 + 105 + 425 + 1691 = 2252;
A000172(6) = 1 + 6 + 33 + 152 + 660 + 2772 + 11560 = 15184; ...

Links

Crossrefs

Cf. A232606, A232683, A232690, A000172.

Programs

  • Mathematica
    Franel[n_] := Sum[Binomial[n, k]^3, {k, 0, n}];
a[0] = 1; a[n_] := Module[{B, G}, B = Sum[Franel[k]*x^k, {k, 0, n+1}] + x^3*O[x]^n; G = 1+x*O[x]^n; For[i=1, i <= n, i++, G = 1+Integrate[(B-1)* (G/x)-B*G^2, x]]; SeriesCoefficient[x/InverseSeries[x*G, x], {x, 0, n}]];
Table[a[n], {n, 0, 26}] (* Jean-François Alcover, Jan 15 2018, translated from 2nd PARI program *)
  • PARI
    /* By Definition (slow): */
{Franel(n)=sum(k=0,n,binomial(n,k)^3)}
{a(n)=if(n==0, 1, (Franel(n) - sum(k=0, n, polcoeff(sum(j=0, min(k, n-1), a(j)*x^j)^n + x*O(x^k), k)))/n)}
for(n=0, 20, print1(a(n), ", "))
  • PARI
    /* Faster, using series reversion: */
{Franel(n)=sum(k=0,n,binomial(n,k)^3)}
{a(n)=local(B=sum(k=0, n+1, Franel(k)*x^k)+x^3*O(x^n), G=1+x*O(x^n));
for(i=1, n, G = 1 + intformal( (B-1)*G/x - B*G^2)); polcoeff(x/serreverse(x*G), n)}
for(n=0, 30, print1(a(n), ", "))

Formula

Given g.f. A(x), Sum_{k=0..n} [x^k] A(x)^n = Sum_{k=0..n} C(n,k)^3 = A000172(n).
Given g.f. A(x), let G(x) = A(x*G(x)) then (G(x) + x*G'(x)) / (G(x) - x*G(x)^2) = Sum_{n>=0} (3*n)!/n!^3 * x^(2*n)/(1-2*x)^(3*n+1) = Sum_{n>=0} A000172(n)*x^n.

A244650 G.f. A(x) satisfies: Sum_{k=0..n} [x^k] A(x)^n = binomial(5*n,2*n).

Original entry on oeis.org

1, 9, 55, 290, 1430, 6827, 32083, 149665, 696130, 3236140, 15055750, 70149880, 327464665, 1531766715, 7180234915, 33728718755, 158765477150, 748819793990, 3538574254840, 16752185111615, 79445373603241, 377382842713434, 1795459769465370, 8554888685073420, 40819261337588995
Offset: 0

Views

Author

Paul D. Hanna, Jul 03 2014

Keywords

Comments

Compare to Sum_{k=0..n} [x^k] 1/(1-x)^(4*n) = binomial(5*n,n).
Compare to Sum_{k=0..n} [x^k] 1/((1-x)*(1-2*x)^2)^n = binomial(4*n,2*n).

Examples

			G.f.: A(x) = 1 + 9*x + 55*x^2 + 290*x^3 + 1430*x^4 + 6827*x^5 +...
ILLUSTRATION OF INITIAL TERMS.
If we form an array of coefficients of x^k in A(x)^n, n>=0, like so:
A^0: [1], 0,    0,     0,      0,        0,         0, ...;
A^1: [1,  9],  55,   290,   1430,     6827,     32083, ...;
A^2: [1, 18,  191], 1570,  11105,    71294,    428452, ...;
A^3: [1, 27,  408,  4569], 42390,   345546,   2564272, ...;
A^4: [1, 36,  706, 10016, 115211], 1142108,  10130498, ...;
A^5: [1, 45, 1085, 18640, 256055,  2992934], 30938150, ...;
A^6: [1, 54, 1545, 31170, 497970,  6708456,  79254029],...; ...
then we can illustrate how the sum of the coefficients of x^k, k=0..n, in A(x)^n (shown above in brackets) equals C(5*n,2*n):
C( 0, 0) = 1 = 1;
C( 5, 2) = 1 +  9 = 10;
C(10, 4) = 1 + 18 +  191 = 210;
C(15, 6) = 1 + 27 +  408 +  4569 = 5005;
C(20, 8) = 1 + 36 +  706 + 10016 +  115211 = 125970;
C(25,10) = 1 + 45 + 1085 + 18640 +  256055 +  2992934 = 3268760;
C(30,12) = 1 + 54 + 1545 + 31170 +  497970 +  6708456 +  79254029 = 86493225; ...

Links

Crossrefs

Cf. A232606, A232683, A232687, A244650, A244651, A244652, A244653, A244654.

Programs

  • PARI
    /* By Definition (slow): */
{a(n)=if(n==0, 1, ( binomial(5*n,2*n) - sum(k=0, n, polcoeff(sum(j=0, min(k, n-1), a(j)*x^j/1!)^n + x*O(x^k), k)))/n)}
for(n=0, 20, print1(a(n), ", "))
  • PARI
    /* Faster, using series reversion: */
{a(n)=local(B=sum(k=0, n+1, binomial(5*k,2*k)*x^k)+x^3*O(x^n), G=1+x*O(x^n));
for(i=1, n, G = 1 + intformal( (B-1)*G/x - B*G^2)); polcoeff(x/serreverse(x*G), n)}
for(n=0, 30, print1(a(n), ", "))

Formula

Recurrence: n*(n+5)*a(n) = (n+1)*(7*n+20)*a(n-1) - (n+2)*(11*n+15)*a(n-2) + 5*(n+1)*(n+2)*a(n-3). - Vaclav Kotesovec, Jul 04 2014
a(n) ~ 5^(n+11/2) / (32*sqrt(Pi)*n^(3/2)). - Vaclav Kotesovec, Jul 04 2014

A244651 G.f. A(x) satisfies: Sum_{k=0..n} [x^k] A(x)^n = binomial(6*n,2*n).

Original entry on oeis.org

1, 14, 135, 1148, 9325, 74634, 596083, 4775288, 38447961, 311305350, 2534757855, 20749571316, 170705908421, 1410874891522, 11710273480395, 97573698950384, 815919118022833, 6845174820882174, 57601263531202871, 486057767175907180, 4112073577799441181, 34871360280503319674
Offset: 0

Views

Author

Paul D. Hanna, Jul 03 2014

Keywords

Comments

Compare to Sum_{k=0..n} [x^k] 1/(1-x)^(5*n) = binomial(6*n,n).
Compare to Sum_{k=0..n} [x^k] 1/((1-x)*(1-2*x)^2)^n = binomial(4*n,2*n).

Examples

			G.f.: A(x) = 1 + 14*x + 135*x^2 + 1148*x^3 + 9325*x^4 + 74634*x^5 +...
ILLUSTRATION OF INITIAL TERMS.
If we form an array of coefficients of x^k in A(x)^n, n>=0, like so:
A^0: [1], 0,    0,      0,       0,        0,          0, ...;
A^1: [1, 14], 135,   1148,    9325,    74634,     596083, ...;
A^2: [1, 28,  466],  6076,   69019,   720328,    7117572, ...;
A^3: [1, 42,  993,  17528], 258462,  3377556,   40526262, ...;
A^4: [1, 56, 1716,  38248,  695450, 10968552,  155816996, ...;
A^5: [1, 70, 2635,  70980, 1536195, 28435134,  467948465, ...;
A^6: [1, 84, 3750, 118468, 2975325, 63276528, 1185303544],...; ...
then we can illustrate how the sum of the coefficients of x^k, k=0..n, in A(x)^n (shown above in brackets) equals C(6*n,2*n):
C( 0, 0) = 1 = 1;
C( 6, 2) = 1 + 14 = 15;
C(12, 4) = 1 + 28 +  466 = 495;
C(18, 6) = 1 + 42 +  993 +  17528 = 18564;
C(24, 8) = 1 + 56 + 1716 +  38248 +  695450 = 735471;
C(30,10) = 1 + 70 + 2635 +  70980 + 1536195 + 28435134 = 30045015;
C(36,12) = 1 + 84 + 3750 + 118468 + 2975325 + 63276528 + 1185303544 = 1251677700; ...

Links

Crossrefs

Cf. A232606, A232683, A232687, A244650, A244652, A244653, A244654.

Programs

  • PARI
    /* By Definition (slow): */
{a(n)=if(n==0, 1, ( binomial(6*n,2*n) - sum(k=0, n, polcoeff(sum(j=0, min(k, n-1), a(j)*x^j/1!)^n + x*O(x^k), k)))/n)}
for(n=0, 20, print1(a(n), ", "))
  • PARI
    /* Faster, using series reversion: */
{a(n)=local(B=sum(k=0, n+1, binomial(6*k,2*k)*x^k)+x^3*O(x^n), G=1+x*O(x^n));
for(i=1, n, G = 1 + intformal( (B-1)*G/x - B*G^2)); polcoeff(x/serreverse(x*G), n)}
for(n=0, 30, print1(a(n), ", "))

Formula

Recurrence: n*(n+3)*(4*n+1)*a(n) = (44*n^3 + 91*n^2 + 99*n + 46)*a(n-1) - (76*n^3 + 159*n^2 + 53*n - 60)*a(n-2) + 9*(n-1)*(n+2)*(4*n+5)*a(n-3). - Vaclav Kotesovec, Jul 04 2014
a(n) ~ 9^(n+4) / (64*sqrt(2*Pi)*n^(3/2)). - Vaclav Kotesovec, Jul 04 2014

A244652 G.f. A(x) satisfies: Sum_{k=0..n} [x^k] A(x)^n = binomial(6*n,3*n).

Original entry on oeis.org

1, 19, 262, 3322, 41455, 520165, 6602716, 84860884, 1103478733, 14500102087, 192309166018, 2571407785918, 34631087423419, 469382779109305, 6398055968407480, 87653105740545976, 1206315271455768505, 16669999282643795899, 231219555870655381438, 3217973871571202211778
Offset: 0

Views

Author

Paul D. Hanna, Jul 03 2014

Keywords

Comments

Compare to Sum_{k=0..n} [x^k] 1/(1-x)^(5*n) = binomial(6*n,n).
Compare to Sum_{k=0..n} [x^k] 1/((1-x)*(1-2*x)^2)^n = binomial(4*n,2*n).

Examples

			G.f.: A(x) = 1 + 19*x + 262*x^2 + 3322*x^3 + 41455*x^4 + 520165*x^5 +...
ILLUSTRATION OF INITIAL TERMS.
If we form an array of coefficients of x^k in A(x)^n, n>=0, like so:
A^0: [1],  0,    0,      0,        0,         0,          0, ...;
A^1: [1,  19], 262,   3322,    41455,    520165,    6602716, ...;
A^2: [1,  38,  885], 16600,   277790,   4356348,   65729806, ...;
A^3: [1,  57, 1869,  46693],  992751,  19018983,  339483259, ...;
A^4: [1,  76, 3214, 100460,  2600405], 59206736, 1229790360, ...;
A^5: [1,  95, 4920, 184760,  5645140, 149282604],3549124200, ...;
A^6: [1, 114, 6987, 306452, 10801665, 325750014, 8738270067],...; ...
then we can illustrate how the sum of the coefficients of x^k, k=0..n, in A(x)^n (shown above in brackets) equals C(6*n,3*n):
C( 0, 0) = 1 = 1;
C( 6, 3) = 1 +  19 = 20;
C(12, 6) = 1 +  38 +  885 = 924;
C(18, 9) = 1 +  57 + 1869 +  46693 = 48620;
C(24,18) = 1 +  76 + 3214 + 100460 +  2600405 = 2704156;
C(30,21) = 1 +  95 + 4920 + 184760 +  5645140 + 149282604 = 155117520;
C(36,24) = 1 + 114 + 6987 + 306452 + 10801665 + 325750014 + 8738270067 = 9075135300; ...

Links

Crossrefs

Cf. A232606, A232683, A232687, A244650, A244651, A244653, A244654.

Programs

  • PARI
    /* By Definition (slow): */
{a(n)=if(n==0, 1, ( binomial(6*n,3*n) - sum(k=0, n, polcoeff(sum(j=0, min(k, n-1), a(j)*x^j/1!)^n + x*O(x^k), k)))/n)}
for(n=0, 20, print1(a(n), ", "))
  • PARI
    /* Faster, using series reversion: */
{a(n)=local(B=sum(k=0, n+1, binomial(6*k,3*k)*x^k)+x^3*O(x^n), G=1+x*O(x^n));
for(i=1, n, G = 1 + intformal( (B-1)*G/x - B*G^2)); polcoeff(x/serreverse(x*G), n)}
for(n=0, 30, print1(a(n), ", "))

Formula

Recurrence: n*(n+2)*(n^2 + 5*n - 2)*a(n) = (18*n^4 + 103*n^3 + 39*n^2 + 56*n + 12)*a(n-1) - (49*n^4 + 264*n^3 - 169*n^2 - 372*n - 180)*a(n-2) + 8*(6*n^4 + 31*n^3 - 51*n^2 - 112*n - 48)*a(n-3) - 16*(n-3)*(n+1)*(n^2 + 7*n + 4)*a(n-4). - Vaclav Kotesovec, Jul 04 2014
a(n) ~ 64*sqrt(6*(26*sqrt(3)-45)) * (8+4*sqrt(3))^n / (sqrt(Pi)*n^(3/2)). - Vaclav Kotesovec, Jul 04 2014

A244653 G.f. A(x) satisfies: Sum_{k=0..n} [x^k] A(x)^n = binomial(7*n,2*n).

Original entry on oeis.org

1, 20, 280, 3521, 42945, 521913, 6379233, 78640740, 978172724, 12270946072, 155139813381, 1975245161155, 25308115539770, 326106155857041, 4223481710794292, 54951230993010196, 717942326681863941, 9415448193554916520, 123904268078599269723, 1635676807214777434793
Offset: 0

Views

Author

Paul D. Hanna, Jul 03 2014

Keywords

Comments

Compare to Sum_{k=0..n} [x^k] 1/(1-x)^(6*n) = binomial(7*n,n).
Compare to Sum_{k=0..n} [x^k] 1/((1-x)*(1-2*x)^2)^n = binomial(4*n,2*n).

Examples

			G.f.: A(x) = 1 + 20*x + 280*x^2 + 3521*x^3 + 42945*x^4 + 521913*x^5 +...
ILLUSTRATION OF INITIAL TERMS.
If we form an array of coefficients of x^k in A(x)^n, n>=0, like so:
A^0: [1],  0,    0,      0,        0,         0,           0, ...;
A^1: [1,  20], 280,   3521,    42945,    521913,     6379233, ...;
A^2: [1,  40,  960], 18242,   305130,   4733386,    70081627, ...;
A^3: [1,  60, 2040,  52163], 1122555,  21563619,   382898782, ...;
A^4: [1,  80, 3520, 113284,  2991220], 68901812,  1437454298, ...;
A^5: [1, 100, 5400, 209605,  6567125, 176797165], 4251203775, ...;
A^6: [1, 120, 7680, 349126, 12666270, 390658878, 10654434813],...; ...
then we can illustrate how the sum of the coefficients of x^k, k=0..n, in A(x)^n (shown above in brackets) equals C(7*n,2*n):
C( 0, 0) = 1 = 1;
C( 7, 2) = 1 + 20 = 21;
C(14, 4) = 1 +  40 +  960 = 1001;
C(21, 6) = 1 +  60 + 2040 +  52163 = 54264;
C(28, 8) = 1 +  80 + 3520 + 113284 +  2991220 = 3108105;
C(35,10) = 1 + 100 + 5400 + 209605 +  6567125 + 176797165 = 183579396;
C(42,12) = 1 + 120 + 7680 + 349126 + 12666270 + 390658878 + 10654434813 = 11058116888; ...

Links

Crossrefs

Cf. A232606, A232683, A232687, A244650, A244651, A244652, A244654.

Programs

  • PARI
    /* By Definition (slow): */
{a(n)=if(n==0, 1, ( binomial(7*n,2*n) - sum(k=0, n, polcoeff(sum(j=0, min(k, n-1), a(j)*x^j/1!)^n + x*O(x^k), k)))/n)}
for(n=0, 20, print1(a(n), ", "))
  • PARI
    /* Faster, using series reversion: */
{a(n)=local(B=sum(k=0, n+1, binomial(7*k,2*k)*x^k)+x^3*O(x^n), G=1+x*O(x^n));
for(i=1, n, G = 1 + intformal( (B-1)*G/x - B*G^2)); polcoeff(x/serreverse(x*G), n)}
for(n=0, 30, print1(a(n), ", "))

Formula

Recurrence: 3*n*(n+1)*(3*n+5)*(3*n+7)*(325*n^3 - 539*n^2 + 200*n - 28)*a(n) = 6*n*(26325*n^6 + 52866*n^5 - 3170*n^4 - 58740*n^3 - 48103*n^2 - 27502*n - 8876)*a(n-1) - (528125*n^7 + 627900*n^6 - 1376202*n^5 - 1623792*n^4 - 330807*n^3 + 114228*n^2 + 109732*n - 4704)*a(n-2) + 4*(177125*n^7 + 79020*n^6 - 878777*n^5 - 465945*n^4 + 871822*n^3 + 881769*n^2 + 181790*n - 53508)*a(n-3) - (n+2)*(404625*n^6 - 896280*n^5 - 811152*n^4 + 2278486*n^3 + 2725599*n^2 + 552482*n - 206976)*a(n-4) + 14*(n+1)*(n+2)*(4225*n^5 - 14157*n^4 + 29689*n^3 + 70969*n^2 + 21750*n - 7308)*a(n-5) + 49*(n-7)*n*(n+1)*(n+2)*(325*n^3 + 436*n^2 + 97*n - 42)*a(n-6). - Vaclav Kotesovec, Jul 04 2014
a(n) ~ c * (7+14/9*sqrt(21))^n / (sqrt(Pi)*n^(3/2)), where c = 43.267577625713256769244376361089321461925061695487162410160820989... . - Vaclav Kotesovec, Jul 04 2014

A244654 G.f. A(x) satisfies: Sum_{k=0..n} [x^k] A(x)^n = binomial(7*n,3*n).

Original entry on oeis.org

1, 34, 889, 22344, 568750, 14812084, 394432598, 10708188328, 295488284471, 8266624187654, 233974149056711, 6688412821905136, 192840384283521996, 5601534217892577384, 163776154208030704124, 4816121399286395128048, 142353930553713780303773, 4226997830260963262597162
Offset: 0

Views

Author

Paul D. Hanna, Jul 03 2014

Keywords

Comments

Compare to Sum_{k=0..n} [x^k] 1/(1-x)^(6*n) = binomial(7*n,n).
Compare to Sum_{k=0..n} [x^k] 1/((1-x)*(1-2*x)^2)^n = binomial(4*n,2*n).

Examples

			G.f.: A(x) = 1 + 34*x + 889*x^2 + 22344*x^3 + 568750*x^4 + 14812084*x^5 +...
ILLUSTRATION OF INITIAL TERMS.
If we form an array of coefficients of x^k in A(x)^n, n>=0, like so:
A^0: [1],  0,     0,       0,        0,           0, ...;
A^1: [1,  34],  889,   22344,   568750,    14812084, ...;
A^2: [1,  68,  2934], 105140,  3447213,   108026800, ...;
A^3: [1, 102,  6135, 287692], 11718441,   437745882, ...;
A^4: [1, 136, 10492,  609304, 29801822], 1301836088, ...;
A^5: [1, 170, 16005, 1109280, 63453080,  3183364624],...; ...
then we can illustrate how the sum of the coefficients of x^k, k=0..n, in A(x)^n (shown above in brackets) equals C(7*n,3*n):
C( 0, 0) = 1 = 1;
C( 7, 3) = 1 +  34 = 35;
C(14, 6) = 1 +  68 +  2934 = 3003;
C(21, 9) = 1 + 102 +  6135 + 287692 = 293930;
C(28,12) = 1 + 136 + 10492 +  609304 + 29801822 = 30421755;
C(35,15) = 1 + 170 + 16005 + 1109280 + 63453080 +  3183364624 = 3247943160; ...

Links

Crossrefs

Cf. A232606, A232683, A232687, A244650, A244651, A244652, A244653.

Programs

  • PARI
    /* By Definition (slow): */
{a(n)=if(n==0, 1, ( binomial(7*n,3*n) - sum(k=0, n, polcoeff(sum(j=0, min(k, n-1), a(j)*x^j/1!)^n + x*O(x^k), k)))/n)}
for(n=0, 20, print1(a(n), ", "))
  • PARI
    /* Faster, using series reversion: */
{a(n)=local(B=sum(k=0, n+1, binomial(7*k,3*k)*x^k)+x^3*O(x^n), G=1+x*O(x^n));
for(i=1, n, G = 1 + intformal( (B-1)*G/x - B*G^2)); polcoeff(x/serreverse(x*G), n)}
for(n=0, 30, print1(a(n), ", "))

Formula

a(n) ~ c * d^n / (sqrt(Pi) * n^(3/2)), where d = 32.201406653616068490560634175718122449630172934... is the root of the equation 67228 - 48020*d - 199969*d^2 + 287875*d^3 - 109375*d^4 + 3125*d^5 = 0, and c = 14.332013639348773543921130720591338... . - Vaclav Kotesovec, Jul 04 2014

A244577 G.f. A(x) satisfies the property that the sum of the coefficients of x^k, k=0..n, in A(x)^n equals (n+1)!.

Original entry on oeis.org

1, 1, 2, 14, 196, 4652, 166168, 8232296, 535974416, 44186331248, 4489336764064, 550549455440096, 80153857492836928, 13665883723351362752, 2697370187692768024448, 610301579538939633684608, 156933087218604923576672512, 45515622704384079509089136384, 14789652457653705738777659937280
Offset: 0

Views

Author

Paul D. Hanna, Jun 30 2014

Keywords

Examples

			E.g.f.: A(x) = 1 + x + 2*x^2/2! + 14*x^3/3! + 196*x^4/4! + 4652*x^5/5! +...
ILLUSTRATION OF INITIAL TERMS.
If we form an array of coefficients of x^k/k! in A(x)^n, n>=0, like so:
A^0: [1],0,  0,   0,     0,      0,       0,         0,           0, ...;
A^1: [1, 1], 2,  14,   196,   4652,  166168,   8232296,   535974416, ...;
A^2: [1, 2,  6], 40,   528,  11824,  403840,  19373792,  1232259840, ...;
A^3: [1, 3, 12,  84], 1068,  22716,  741456,  34375200,  2132407248, ...;
A^4: [1, 4, 20, 152,  1912], 39008, 1218496,  54513152,  3292657664, ...;
A^5: [1, 5, 30, 250,  3180,  62980],1889080,  81499400,  4785873360, ...;
A^6: [1, 6, 42, 384,  5016,  97632, 2826288],117620256,  6706638336, ...;
A^7: [1, 7, 56, 560,  7588, 146804, 4127200, 165911312], 9177810320, ...;
A^8: [1, 8, 72, 784, 11088, 215296, 5918656, 230372480, 12358846848], ...; ...
then we can illustrate how the sum of the coefficients of x^k, k=0..n, in A(x)^n (shown above in brackets) equals (n+1)!:
1! = 1;
2! = 1 + 1;
3! = 1 + 2 + 6/2!;
4! = 1 + 3 + 12/2! + 84/3!;
5! = 1 + 4 + 20/2! + 152/3! + 1912/4!;
6! = 1 + 5 + 30/2! + 250/3! + 3180/4! + 62980/5!; ...

Links

Crossrefs

Cf. A244589, A232606, A232607, A232683, A232687.

Programs

  • PARI
    /* By Definition (slow): */
{a(n)=if(n==0, 1, n!*((n+1)! - sum(k=0, n, polcoeff(sum(j=0, min(k, n-1), a(j)*x^j/j!)^n + x*O(x^k), k)))/n)}
for(n=0, 20, print1(a(n), ", "))
  • PARI
    /* Faster, using series reversion: */
{a(n)=local(B=sum(k=0, n+1, (k+1)!*x^k)+x^3*O(x^n), G=1+x*O(x^n));
for(i=1, n, G = 1 + intformal( (B-1)*G/x - B*G^2)); n!*polcoeff(x/serreverse(x*G), n)}
for(n=0, 30, print1(a(n), ", "))

Formula

Given g.f. A(x), Sum_{k=0..n} [x^k] A(x)^n = (n+1)!.
a(n) ~ exp(-1) * (n!)^2. - Vaclav Kotesovec, Jul 03 2014
Showing 1-8 of 8 results.

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