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 1-4 of 4 results.

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
Showing 1-4 of 4 results.

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