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 1-6 of 6 results.

A171791 G.f. A(x) satisfies: [x^n] A(x)^((n+1)^2) = 0 for n>1 with a(0)=a(1)=1.

Original entry on oeis.org

1, 1, -4, 25, -194, 1603, -15264, 122316, -1897710, -8845133, -1169435932, -52853978047, -3193246498792, -205347570309000, -14534295599537024, -1115833257773950536, -92445637289048967654, -8219735646409095418617
Offset: 0

Views

Author

Paul D. Hanna, Jan 24 2010

Keywords

Comments

It appears that for k>0, a(k) is odd iff k = 2*A003714(n)+1 for n>=0, where A003714 is the fibbinary numbers (integers whose binary representation contains no consecutive ones); this is true for at least the first 520 terms. [See also A263190 and A263075.] - Paul D. Hanna, Oct 09 2013
Observation of Paul D. Hanna is true for at least the first 1028 terms. - Sean A. Irvine, Apr 25 2014

Examples

			G.f.: A(x) = 1 + x - 4*x^2 + 25*x^3 - 194*x^4 + 1603*x^5 +...
The coefficients in the square powers of g.f. A(x) begin:
A^1:  [1,  1,   -4,    25,   -194,    1603,   -15264,    122316, ...];
A^4:  [1,  4,  -10,    56,   -427,    3360,   -33546,    218880, ...];
A^9:  [1,  9,    0,    21,   -252,    1701,   -25992,     -2970, ...];
A^16: [1, 16,   56,     0,    -84,    -784,   -18656,   -384896, ...];
A^25: [1, 25,  200,   525,      0,   -2695,   -38600,   -878150, ...];
A^36: [1, 36,  486,  3000,   7821,       0,  -101322,  -1916352, ...];
A^49: [1, 49,  980, 10241,  58898,  170079,        0,  -4515000, ...];
A^64: [1, 64, 1760, 27136, 256048, 1500352,  4979712,         0, ...];
A^81: [1, 81, 2916, 61425, 838026, 7720839, 48097152, 184870512, 0,...]; ...
Note how the coefficient of x^n in A(x)^((n+1)^2) = 0 for n>1.
ALTERNATE RELATION.
The coefficients in A(x)^(n^2) * (1 - n*x*A(x)'/A(x)) begin:
n=1: [1, 0, 4, -50, 582, -6412, 76320, -733896, 13283970, ...];
n=2: [1, 2, 0, -28, 427, -5040, 67092, -547200, 15539502, ...];
n=3: [1, 6, 0, 0, 84, -1134, 25992, 3960, 13172355, ...];
n=4: [1, 12, 28, 0, 0, 196, 9328, 288672, 13426530, ...];
n=5: [1, 20, 120, 210, 0, 0, 7720, 351260, 15775425, ...];
n=6: [1, 30, 324, 1500, 2607, 0, 0, 319392, 17452530, ...];
n=7: [1, 42, 700, 5852, 25242, 48594, 0, 0, 15518020, ...];
n=8: [1, 56, 1320, 16960, 128024, 562632, 1244928, 0, 0, ...];
n=9: [1, 72, 2268, 40950, 465570, 3431484, 16032384, 41082336, 0, 0, ...]; ...
in which the two adjacent diagonals above the main diagonal are all zeros after initial terms, illustrating that
(1) 0 = [x^(n-1)] A(x)^(n^2) * (1 - n*x*A(x)'/A(x)), and
(2) 0 = [x^n] A(x)^(n^2) * (1 - n*x*A(x)'/A(x)), for n > 0.
		

Crossrefs

Programs

  • PARI
    {a(n) = local(A=[1,1]); for(m=3,n+1, A=concat(A,0); A[ #A]=-Vec(Ser(A)^(m^2))[m]/m^2); A[n+1]}
    for(n=0, 20, print1(a(n), ", "))

Formula

The g.f. A(x) satisfies the following relations.
(1) 0 = [x^(n-1)] A(x)^(n^2), for n > 1.
(2) 0 = [x^(n-1)] A(x)^(n^2) * (1 - n*x*A(x)'/A(x)), for n > 1. - Paul D. Hanna, Oct 22 2020
(3) 0 = [x^n] A(x)^(n^2) * (1 - n*x*A(x)'/A(x)), for n > 0. - Paul D. Hanna, Oct 22 2020

A118113 Even Fibbinary numbers + 1; also 2*Fibbinary(n) + 1.

Original entry on oeis.org

1, 3, 5, 9, 11, 17, 19, 21, 33, 35, 37, 41, 43, 65, 67, 69, 73, 75, 81, 83, 85, 129, 131, 133, 137, 139, 145, 147, 149, 161, 163, 165, 169, 171, 257, 259, 261, 265, 267, 273, 275, 277, 289, 291, 293, 297, 299, 321, 323, 325, 329, 331, 337, 339, 341, 513, 515, 517
Offset: 0

Views

Author

Labos Elemer, Apr 13 2006

Keywords

Comments

m for which binomial(3*m-2,m) (see A117671) is odd, since by Kummer's theorem that happens exactly when the binary expansions of m and 2*m-2 have no 1 bit at the same position in each, and so m odd and no 11 bit pairs except optionally the least significant 2 bits. - Kevin Ryde, Jun 14 2025

Crossrefs

Cf. A003714 (Fibbinary numbers), A022340 (even Fibbinary numbers).

Programs

  • Maple
    F:= combinat[fibonacci]:
    b:= proc(n) local j;
          if n=0 then 0
        else for j from 2 while F(j+1)<=n do od;
             b(n-F(j))+2^(j-2)
          fi
        end:
    a:= n-> 2*b(n)+1:
    seq(a(n), n=0..70);  # Alois P. Heinz, Aug 03 2012
  • Mathematica
    Select[Table[Mod[Binomial[3*k,k], k+1], {k,1200}], #>0&]

Formula

a(n) = A022340(n) + 1.
a(n) = 2*A003714(n) + 1.
Solutions to {x : binomial(3x,x) mod (x+1) != 0 } are given in A022341. The corresponding values of binomial(3x,x) mod (x+1) are given here.

Extensions

New definition from T. D. Noe, Dec 19 2006

A263190 G.f. satisfies: [x^(n-1)] A(x)^(n^2) = (n!)^2 for n>=1.

Original entry on oeis.org

1, 1, 0, 1, 46, 1723, 81104, 4793304, 349869074, 31080492631, 3318717525832, 421195540320465, 62871475566985208, 10927921934497456588, 2191068343727736627744, 502384409006686040020572, 130687814451798554601790746, 38294333521028379285810681487, 12557951067433973525611840784048, 4581888866092825667058378205370595
Offset: 0

Views

Author

Paul D. Hanna, Oct 12 2015

Keywords

Comments

CONJECTURES.
(1) Limit a(n)/(n!)^2 = 1/exp(1).
(2) There are no negative terms.
(3) ODD TERMS: It appears that for k>0, a(k) is odd iff k = 2*A003714(n)+1 for n>=0, where A003714 is the fibbinary numbers (integers whose binary representation contains no consecutive ones); this is true for at least the first 531 terms. [See also A171791 and A263075.]
Conjectures hold to at least a(1000). - Sean A. Irvine, Oct 21 2015

Examples

			G.f.: A(x) = 1 + x + x^3 + 46*x^4 + 1723*x^5 + 81104*x^6 + 4793304*x^7 +...
The coefficients in A(x)^(n^2) begin:
n=1: [1, 1, 0, 1, 46, 1723, 81104, 4793304, 349869074, ...];
n=2: [1, 4, 6, 8, 197, 7456, 345654, 20167888, 1458010566, ...];
n=3: [1, 9, 36, 93, 612, 19197, 866208, 49440834, 3515499819, ...];
n=4: [1, 16, 120, 576, 2796, 44656, 1803872, 99433344, ...];
n=5: [1, 25, 300, 2325, 14400, 130705, 3606800, 183492150, ...];
n=6: [1, 36, 630, 7176, 61821, 518400, 8260086, 332807184, ...];
n=7: [1, 49, 1176, 18473, 216482, 2154775, 25401600, 655445812, ...];
n=8: [1, 64, 2016, 41728, 642352, 8045248, 95405312, 1625702400, ...];
n=9: [1, 81, 3240, 85401, 1673946, 26315199, 360707040, 5266837404, 131681894400, ...]; ...
where the terms along the main diagonal begin:
[1, 4, 36, 576, 14400, 518400, 25401600, 1625702400, ..., (n!)^2, ...].
LOCATION OF ODD TERMS.
Note that odd terms a(n) occur at positions n starting with:
[0, 1, 3, 5, 9, 11, 17, 19, 21, 33, 35, 37, 41, 43, 65, 67, 69, 73, 75, 81, 83, 85, 129, 131, 133, 137, 139, 145, 147, 149, 161, 163, 165, 169, 171, 257, ...],
which seems to equal A118113, the even fibbinary numbers + 1, with an initial zero included.
		

Crossrefs

Programs

  • PARI
    {a(n) = local(A=[1, 1]); for(i=1, n+1, A=concat(A, 0); m=#A; A[m] = ( m!^2 - Vec(Ser(A)^(m^2))[m] )/m^2 ); A[n+1]}
    for(n=0, 20, print1(a(n), ", "))

A244589 E.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)^n.

Original entry on oeis.org

1, 1, 5, 67, 1937, 98791, 7744549, 857382695, 126889656641, 24157912257775, 5749369223697701, 1672527291075462559, 584038879457972531185, 241150002566590866157943, 116245385996298375640197893, 64707252902905394310560934391, 41198982747438307655532993553409
Offset: 0

Views

Author

Paul D. Hanna, Jun 30 2014

Keywords

Examples

			E.g.f.: A(x) = 1 + x + 5*x^2/2! + 67*x^3/3! + 1937*x^4/4! + 98791*x^5/5! +...
where
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, ...;
A^1: [1, 1], 5,   67,  1937,   98791,   7744549,   857382695, ...;
A^2: [1, 2, 12], 164,  4560,  223652,  17054920,  1853019716, ...;
A^3: [1, 3, 21,  297], 8049,  380853,  28237293,  3008400909, ...;
A^4: [1, 4, 32,  472, 12608], 577864,  41657008,  4348646600, ...;
A^5: [1, 5, 45,  695, 18465,  823475], 57747565,  5903103995, ...;
A^6: [1, 6, 60,  972, 25872, 1127916,  77020344], 7706019180, ...;
A^7: [1, 7, 77, 1309, 35105, 1502977, 100075045,  9797289761], ...; ...
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)^n:
1^0 = 1;
2^1 = 1 + 1 = 2;
3^2 = 1 + 2 + 12/2! = 9;
4^3 = 1 + 3 + 21/2! + 297/3! = 64;
5^4 = 1 + 4 + 32/2! + 472/3! + 12608/4! = 625;
6^5 = 1 + 5 + 45/2! + 695/3! + 18465/4! + 823475/5! = 7776;
7^6 = 1 + 6 + 60/2! + 972/3! + 25872/4! + 1127916/5! + 77020344/6! = 117649; ...
		

Crossrefs

Programs

  • PARI
    /* By Definition (slow): */
    {a(n)=if(n==0, 1, n!*((n+1)^n - 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)^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)); n!*polcoeff(x/serreverse(x*G), n)}
    for(n=0, 30, print1(a(n), ", "))

Formula

E.g.f. A(x) satisfies: Sum_{k=0..n} [x^k] A(x)^n = (n+1)^n.
a(n) ~ exp(1-exp(-1)) * n! * n^(n-1). - Vaclav Kotesovec, Jul 03 2014

A294360 G.f. A(x) satisfies: [x^(n-1)] A(x)^(n^2) = (n^2)^(n-1) for n>=1.

Original entry on oeis.org

1, 1, 5, 146, 9935, 1161399, 206499453, 52093726159, 17770811461875, 7903030237890371, 4450363873663943294, 3098938855124650814264, 2616552190721485829559668, 2635178871851323631797948230, 3121810359776427044817295874677, 4298670834657263815567279951080956, 6809336162211769799756516349665301635, 12296952422064277377043754761717448273557, 25116528778581121454413639996325045161219974
Offset: 0

Views

Author

Paul D. Hanna, Nov 01 2017

Keywords

Examples

			G.f.: A(x) = 1 + x + 5*x^2 + 146*x^3 + 9935*x^4 + 1161399*x^5 + 206499453*x^6 + 52093726159*x^7 + 17770811461875*x^8 + 7903030237890371*x^9 + 4450363873663943294*x^10 + 3098938855124650814264*x^11 + 2616552190721485829559668*x^12 +...
such that the coefficient of x^(n-1) in A(x)^(n^2) equals (n^2)^(n-1) for n>=1.
The table of coefficients of x^k in A(x)^(n^2) begin:
n=1: [1, 1, 5, 146, 9935, 1161399, 206499453, ...];
n=2: [1, 4, 26, 648, 41703, 4775648, 840796570, ...];
n=3: [1, 9, 81, 1758, 102213, 11266209, 1949437539, ...];
n=4: [1, 16, 200, 4096, 207220, 21470032, 3617873616, ...];
n=5: [1, 25, 425, 8950, 390625, 36920005, 5985228975, ...];
n=6: [1, 36, 810, 18696, 723375, 60466176, 9272944890, ...];
n=7: [1, 49, 1421, 37338, 1347843, 97588547, 13841287201, ...];
n=8: [1, 64, 2336, 71168, 2535248, 159036480, 20303433408, 4398046511104, ...]; ...
in which the main diagonal begins:
[1, 4, 81, 4096, 390625, 60466176, 13841287201, 4398046511104, ..., (n^2)^(n-1), ...].
		

Crossrefs

Programs

  • PARI
    {a(n) = my(A=[1]); for(m=2,n+1, A = concat(A,0); A[m] = ( (m^2)^(m-1) - Vec( Ser(A)^(m^2) )[m] )/m^2);A[n+1]}
    for(n=0,20,print1(a(n),", "))

Formula

a(n) ~ c * n^(2*n - 2), where c = exp(2 - exp(-2)) = 6.453771681742981632532303... - Vaclav Kotesovec, Aug 11 2021, updated Mar 18 2024

A371673 Expansion of g.f. A(x) satisfying [x^(n-1)] A(x)^(n^2) = A000108(n-1) * n^n for n >= 1, where A000108 is the Catalan numbers.

Original entry on oeis.org

1, 1, 2, 15, 284, 8575, 345460, 17190684, 1012901520, 68810750943, 5291667341342, 454479660308531, 43140290728900554, 4487833959824527910, 508072065566891421336, 62222074620010689986918, 8200304581300850453687880, 1157674985567876068399895997, 174357014524193551292388873190
Offset: 0

Views

Author

Paul D. Hanna, Apr 02 2024

Keywords

Comments

Conjecture: a(n) is odd for n > 0 iff n = 2*A003714(k) + 1 for some k, where A003714 is the Fibbinary numbers (integers whose binary representation contains no consecutive ones). See A263075, A263190, and A171791.

Examples

			G.f.: A(x) = 1 + x + 2*x^2 + 15*x^3 + 284*x^4 + 8575*x^5 + 345460*x^6 + 17190684*x^7 + 1012901520*x^8 + 68810750943*x^9 + 5291667341342*x^10 + ...
The table of coefficients of x^k in A(x)^(n^2) begin:
 n=1: [1,  1,    2,    15,    284,    8575,    345460, ...];
 n=2: [1,  4,   14,    88,   1365,   38304,   1497150, ...];
 n=3: [1,  9,   54,   363,   4410,  105705,   3874824, ...];
 n=4: [1, 16,  152,  1280,  13804,  263408,   8535648, ...];
 n=5: [1, 25,  350,  3875,  43750,  688205,  18352800, ...];
 n=6: [1, 36,  702, 10200, 133389, 1959552,  42189822, ...];
 n=7: [1, 49, 1274, 23863, 376320, 5810763, 108707676, ...];
 ...
where the terms along the main diagonal start as
 [1, 4, 54, 1280, 43750, 1959552, 108707676, ...]
which equals A000108(n-1)*n^n for n >= 1:
 [1, 1*2^2, 2*3^3, 5*4^4, 14*5^5, 42*6^6, 132*7^7, ...].
Compare the above table to the coefficients in 1/(1 - n*x)^n:
 n=1: [1,  1,    1,     1,      1,       1,         1, ...];
 n=2: [1,  4,   12,    32,     80,     192,       448, ...];
 n=3: [1,  9,   54,   270,   1215,    5103,     20412, ...];
 n=4: [1, 16,  160,  1280,   8960,   57344,    344064, ...];
 n=5: [1, 25,  375,  4375,  43750,  393750,   3281250, ...];
 n=6: [1, 36,  756, 12096, 163296, 1959552,  21555072, ...];
 n=7: [1, 49, 1372, 28812, 504210, 7764834, 108707676, ...];
 ...
to see that the main diagonals are equal.
		

Crossrefs

Programs

  • PARI
    {a(n) = my(A=[1], m); for(i=1,n, A=concat(A,0); m=#A;
    A[m] = ( m^m*binomial(2*m-1,m-1)/(2*m-1) - Vec( Ser(A)^(m^2) )[m] )/(m^2) );A[n+1]}
    for(n=0,30,print1(a(n),", "))

Formula

G.f. A(x) = Sum_{n>=0} a(n)*x^n satisfies the following formulas.
(1) [x^(n-1)] A(x)^(n^2) = n^n * binomial(2*n-1,n-1)/(2*n-1) for n >= 1.
(2) [x^(n-1)] A(x)^(n^2) = [x^(n-1)] 1/(1 - n*x)^n for n >= 1.
Showing 1-6 of 6 results.