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.

A231619 E.g.f. A(x) satisfies: A'(x) = A(x*A'(x)) with A(0)=1.

Original entry on oeis.org

1, 1, 1, 3, 18, 174, 2394, 43488, 993150, 27560142, 906516252, 34681891932, 1520713138896, 75519376934904, 4206425439922920, 260667523550924424, 17848073907441578484, 1342326288499671643956, 110319548590679184794880, 9862994518551295719972264, 955390360741496204485599576
Offset: 0

Views

Author

Paul D. Hanna, Nov 14 2013

Keywords

Comments

Conjectures:
a(n) == 0 (mod 2) for n>=4.
a(n) == 0 (mod 3) for n>=3.
a(n) == 0 (mod 27) for n>=9.
a(n) == 0 (mod 2^2) for n>=10.
a(n) == 0 (mod 2^3) for n>=18.
a(n) == 0 (mod 2^k) for n>=(8*k-6), k>1.
a(8*n) == 2^n (mod 2^(n+1)) for n>=0.

Examples

			E.g.f.: A(x) = 1 + x + x^2/2! + 3*x^3/3! + 18*x^4/4! + 174*x^5/5! + 2394*x^6/6! +...
such that
A(x*A'(x)) = A'(x) = 1 + x + 3*x^2/2! + 18*x^3/3! + 174*x^4/4! + 2394*x^5/5! +...
To illustrate a(n) = [x^(n-1)/(n-1)!] A(x)^n/n, create a table of coefficients of x^k/k!, k>=0, in A(x)^n like so:
A^1: [1, 1,  1,   3,   18,   174,   2394,    43488,    993150, ...];
A^2: [1, 2,  4,  12,   66,   588,   7596,   131580,   2897316, ...];
A^3: [1, 3,  9,  33,  180,  1512,  18396,   303210,   6418602, ...];
A^4: [1, 4, 16,  72,  420,  3456,  40104,   630072,  12801888, ...];
A^5: [1, 5, 25, 135,  870,  7290,  82350,  1241820,  24234030, ...];
A^6: [1, 6, 36, 228, 1638, 14364, 161604,  2366388,  44519220, ...];
A^7: [1, 7, 49, 357, 2856, 26628, 304416,  4390470,  80167626, ...];
A^8: [1, 8, 64, 528, 4680, 46752, 551376,  7945200, 142078752, ...];
A^9: [1, 9, 81, 747, 7290, 78246, 961794, 14022072, 248041278, ...]; ...
then the diagonal in the above table generates this sequence shift left:
[1/1, 2/2, 9/3, 72/4, 870/5, 14364/6, 304416/7, 7945200/8, 248041278/9, ...].
SUMS OF TERM RESIDUES MODULO 2^n.
Given a(k) == 0 (mod 2^n) for k>=(8*n-6) for n>1, then it is interesting to consider the sums of the residues of all terms modulo 2^n for n>=1.
Let b(n) = Sum_{k>=0} a(k) (mod 2^n) for n>=1, then the sequence {b(n)} begins:
[4, 16, 44, 156, 428, 1068, 2476, 5804, 13484, 29868, 67756, 149676, 354476, 739500, 1558700, 3131564, 7129260, 14993580, 31246508, 68995244, 153929900, ...].

Links

Crossrefs

Cf. A231866, A231899, A232694, A232695, A232696, A232686.

Programs

  • Mathematica
    terms = 20; A[] = 1; Do[A[x] = 1 + Integrate[A[x*A'[x]] + O[x]^j, x] + O[x]^j // Normal, {j, 1, terms}]; CoefficientList[A[x] , x]*Range[0, terms-1]! (* Jean-François Alcover, Jan 15 2018 *)
  • PARI
    {a(n)=local(A=1+x);for(i=1,n,A=1+intformal(subst(A,x,x*A' +x*O(x^n))));n!*polcoeff(A,n)}
for(n=0,25,print1(a(n),", "))
  • PARI
    {a(n)=local(A=1+x);for(i=1,n,A=1+intformal(1/x*serreverse(x/A +x*O(x^n))));n!*polcoeff(A,n)}
for(n=0,25,print1(a(n),", "))

Formula

E.g.f. satisfies: A(x) = A'(x/A(x)).
E.g.f. satisfies: A(x) = x / Series_Reversion( x*A'(x) ).
E.g.f. A(x) satisfies: A''(x) = A'(x) * A'(x*A'(x)) / (1 - x*A'(x*A'(x))).
E.g.f. A(x) satisfies: A''(x/A(x)) = A(x)^2 * A'(x) / (A(x) - x*A'(x)).
a(n) = [x^(n-1)/(n-1)!] A(x)^n/n for n>=1.

A232695 E.g.f. A(x) satisfies: A'(x) = A(x*A'(x)^5) with A(0)=1.

Original entry on oeis.org

1, 1, 1, 11, 266, 10326, 562926, 40058076, 3554828286, 381374161166, 48366170807276, 7128626213386476, 1204840675597360776, 230986547885416953936, 49777541426984300127816, 11964954349177005321013976, 3186498480002528225295506276, 934756070179948684556233837476
Offset: 0

Views

Author

Paul D. Hanna, Nov 27 2013

Keywords

Comments

CONJECTURES.
a(n) == 1 (mod 5) for n>=0.
a(n) == 0 (mod 2) for n>=4.
a(n) == 0 (mod 2^2) for n>=10.
a(n) == 0 (mod 2^3) for n>=18.
a(n) == 0 (mod 2^k) for n>=(8*k-6) for k>1.

Examples

			E.g.f.: A(x) = 1 + x + x^2/2! + 11*x^3/3! + 266*x^4/4! + 10326*x^5/5! +...
such that
A(x*A'(x)^5) = A'(x) = 1 + x + 11*x^2/2! + 266*x^3/3! + 10326*x^4/4! +...
To illustrate a(n) = [x^(n-1)/(n-1)!] A(x)^(5*n-4)/(5*n-4), create a table of coefficients of x^k/k!, k>=0, in A(x)^(5*n-4), n>=1, like so:
A^1 : [1,  1,   1,    11,    266,    10326,     562926, ...];
A^6 : [1,  6,  36,   276,   4086,   124476,    6058956, ...];
A^11: [1, 11, 121,  1441,  21956,   530376,   21460736, ...];
A^16: [1, 16, 256,  4256,  79376,  1891776,   66002016, ...];
A^21: [1, 21, 441,  9471, 216846,  5697426,  191016546, ...];
A^26: [1, 26, 676, 17836, 489866, 14636076,  510313076, ...];
A^31: [1, 31, 961, 30101, 968936, 32971476, 1241800356, ...]; ...
then the diagonal in the above table generates this sequence shift left:
[1/1, 6/6, 121/11, 4256/16, 216846/21, 14636076/26, 1241800356/31, ...].
SUMS OF TERM RESIDUES MODULO 2^n.
Given a(k) == 0 (mod 2^n) for k>=(8*n-6) for n>1, then it is interesting to consider the sums of the residues of all terms modulo 2^n for n>=1.
Let b(n) = Sum_{k>=0} a(k) (mod 2^n) for n>=1, then the sequence {b(n)} begins:
[4, 16, 52, 180, 388, 868, 2532, 5860, 13028, 27364, 63204, 157412, 370404, 780004, 1730276, 3630820, 7431908, 14509796, 32597732, 72967908, ...].

Links

Crossrefs

Cf. A231619, A231866, A231899, A232694, A232696.

Programs

  • PARI
    {a(n)=local(A=1+x); for(i=1, n, A=1+intformal(subst(A, x, x*A'^5 +x*O(x^n)))); n!*polcoeff(A, n)}
for(n=0, 25, print1(a(n), ", "))
  • PARI
    {a(n)=local(A=1+x); for(i=1, n, A=1+intformal((1/x*serreverse(x/A^5 +x*O(x^n)))^(1/5))); n!*polcoeff(A, n)}
for(n=0, 25, print1(a(n), ", "))

Formula

E.g.f. satisfies: A(x) = A'(x/A(x)^5).
E.g.f. satisfies: A(x) = ( x / Series_Reversion( x*A'(x)^5 ) )^(1/5).
a(n) = [x^(n-1)/(n-1)!] A(x)^(5*n-4)/(5*n-4) for n>=1.

A231866 E.g.f. A(x) satisfies: A'(x) = A(x*A'(x)^2) with A(0)=1.

Original entry on oeis.org

1, 1, 1, 5, 53, 909, 22149, 711297, 28687833, 1405408841, 81620841401, 5516637014061, 427699967681709, 37595972586389109, 3711295383595024221, 408142117923542673737, 49663409518409586541937, 6647274714312311181770577, 973638869018128380202018353
Offset: 0

Views

Author

Paul D. Hanna, Nov 14 2013

Keywords

Examples

			E.g.f.: A(x) = 1 + x + x^2/2! + 5*x^3/3! + 53*x^4/4! + 909*x^5/5! + 22149*x^6/6! +...
such that
A(x*A'(x)^2) = A'(x) = 1 + x + 5*x^2/2! + 53*x^3/3! + 909*x^4/4! + 22149*x^5/5! +...
The square of the e.g.f. begins:
A(x)^2 = 1 + 2*x + 4*x^2/2! + 16*x^3/3! + 152*x^4/4! + 2448*x^5/5! +...
To illustrate a(n) = [x^(n-1)/(n-1)!] A(x)^(2*n-1)/(2*n-1), create a table of coefficients of x^k/k!, k>=0, in A(x)^(2*n-1), n>=1, like so:
A^1 : [1,  1,   1,    5,    53,    909,    22149,    711297, ...];
A^3 : [1,  3,   9,   39,   333,   5007,   112101,   3395907, ...];
A^5 : [1,  5,  25,  145,  1205,  16065,   326525,   9235165, ...];
A^7 : [1,  7,  49,  371,  3437,  44163,   825741,  21682143, ...];
A^9 : [1,  9,  81,  765,  8181, 108981,  1952469,  47966553, ...];
A^11: [1, 11, 121, 1375, 16973, 243639,  4370069, 102669787, ...];
A^13: [1, 13, 169, 2249, 31733, 498537,  9246861, 213557877, ...];
A^15: [1, 15, 225, 3435, 54765, 945195, 18486525, 430317495, ...]; ...
then the diagonal in the above table generates this sequence shift left:
[1/1, 3/3, 25/5, 371/7, 8181/9, 243639/11, 9246861/13, 430317495/15, ...].

Crossrefs

Cf. A231619, A231899, A232694, A232695, A232696.

Programs

  • PARI
    {a(n)=local(A=1+x); for(i=1, n, A=1+intformal(subst(A, x, x*A'^2 +x*O(x^n)))); n!*polcoeff(A, n)}
for(n=0, 25, print1(a(n), ", "))
  • PARI
    {a(n)=local(A=1+x);for(i=1,n,A=1+intformal(sqrt(1/x*serreverse(x/A^2 +x*O(x^n)))));n!*polcoeff(A,n)}
for(n=0,25,print1(a(n),", "))

Formula

E.g.f. satisfies: A(x) = A'(x/A(x)^2).
E.g.f. satisfies: A(x) = sqrt( x / Series_Reversion( x*A'(x)^2 ) ).
a(n) = [x^(n-1)/(n-1)!] A(x)^(2*n-1)/(2*n-1) for n>=1.
a(n) == 1 (mod 4) for n>=0.

A232694 E.g.f. A(x) satisfies: A'(x) = A(x*A'(x)^4) with A(0)=1.

Original entry on oeis.org

1, 1, 1, 9, 177, 5601, 249681, 14569545, 1062623265, 93853717761, 9810385567329, 1192614883442889, 166310354311947345, 26308546859152889697, 4677436610087462937393, 927353710845763536487305, 203648424149429271943770945, 49245501579619466882211194625, 13045520297945193508654786790337
Offset: 0

Views

Author

Paul D. Hanna, Nov 27 2013

Keywords

Comments

CONJECTURES.
a(n) == 0 (mod 3) for n>=3.
a(n) == 1 (mod 8) for n>=0.

Examples

			E.g.f.: A(x) = 1 + x + x^2/2! + 9*x^3/3! + 177*x^4/4! + 5601*x^5/5! +...
such that
A(x*A'(x)^4) = A'(x) = 1 + x + 9*x^2/2! + 177*x^3/3! + 5601*x^4/4! +...
To illustrate a(n) = [x^(n-1)/(n-1)!] A(x)^(4*n-3)/(4*n-3), create a table of coefficients of x^k/k!, k>=0, in A(x)^(4*n-3), n>=1, like so:
A^1 : [1,  1,   1,     9,    177,    5601,     249681,    14569545, ...];
A^5 : [1,  5,  25,   165,   2145,   55125,    2211225,   120873045, ...];
A^9 : [1,  9,  81,   801,  10449,  218889,    7501761,   373998465, ...];
A^13: [1, 13, 169,  2301,  35841,  731133,   21950409,   974182989, ...];
A^17: [1, 17, 289,  5049,  95217,  2102577,  60325809,  2417773881, ...];
A^21: [1, 21, 441,  9429, 211617,  5243301, 154446201,  5861076165, ...];
A^25: [1, 25, 625, 15825, 414225, 11585625, 364238625, 13752570225, ...];
A^29: [1, 29, 841, 24621, 738369, 23206989, 791747241, 30816074685, ...]; ...
then the diagonal in the above table generates this sequence shift left:
[1/1, 5/5, 81/9, 2301/13, 95217/17, 5243301/21, 364238625/25, 30816074685/29, ...].

Crossrefs

Cf. A231619, A231866, A231899, A232695, A232696.

Programs

  • PARI
    {a(n)=local(A=1+x); for(i=1, n, A=1+intformal(subst(A, x, x*A'^4 +x*O(x^n)))); n!*polcoeff(A, n)}
for(n=0, 25, print1(a(n), ", "))
  • PARI
    {a(n)=local(A=1+x); for(i=1, n, A=1+intformal((1/x*serreverse(x/A^4 +x*O(x^n)))^(1/4))); n!*polcoeff(A, n)}
for(n=0, 25, print1(a(n), ", "))

Formula

E.g.f. satisfies: A(x) = A'(x/A(x)^4).
E.g.f. satisfies: A(x) = ( x / Series_Reversion( x*A'(x)^4 ) )^(1/4).
a(n) = [x^(n-1)/(n-1)!] A(x)^(4*n-3)/(4*n-3) for n>=1.

A232686 E.g.f. A(x) satisfies: A'(x) = A(x*A'(x))^2.

Original entry on oeis.org

1, 1, 2, 14, 196, 4352, 135864, 5550560, 283148560, 17454554624, 1269351110560, 106925372050688, 10284022937294400, 1116367988620697600, 135485394041604223360, 18237986446534462042112, 2704852975724947252789504, 439428075979518197809971200, 77809530527904746552436748800
Offset: 0

Views

Author

Paul D. Hanna, Dec 03 2013

Keywords

Comments

CONJECTURES.
a(n) == 0 (mod 2) for n>=2.
a(n) == 0 (mod 2^2) for n>=4.
a(n) == 0 (mod 2^3) for n>=5.
a(n) == 0 (mod 2^k) for n>=(2*k-1), k>=3.
a(2*n) == 2^n (mod 2^(n+1)) for n>=0.

Examples

			E.g.f.: A(x) = 1 + x + 2*x^2/2! + 14*x^3/3! + 196*x^4/4! + 4352*x^5/5! +...
such that
A(x*A'(x))^2 = A'(x) = 1 + 2*x + 14*x^2/2! + 196*x^3/3! + 4352*x^4/4! +...
A(x*A'(x)) = sqrt(A'(x)) = 1 + x + 6*x^2/2! + 80*x^3/3! + 1748*x^4/4! + 54392*x^5/5! + 2227608*x^6/6! + 114233232*x^7/7! + 7088554512*x^8/8! +...
To illustrate a(n) = [x^(n-1)/(n-1)!] A(x)^(2*n)/n, create a table of coefficients of x^k/k!, k>=0, in A(x)^(2*n) like so:
A^2: [1,  2,   6,   40,    528,   11224,   339632,   13560864, ...];
A^4: [1,  4,  20,  152,   1912,   37808,  1075680,   40938272, ...];
A^6: [1,  6,  42,  384,   5016,   95832,  2590464,   94032000, ...];
A^8: [1,  8,  72,  784,  11088,  212896,  5575424,  194217024, ...];
A^10:[1, 10, 110, 1400,  21760,  431960, 11203920,  378660320, ...];
A^12:[1, 12, 156, 2280,  39048,  815184, 21350112,  709540704, ...];
A^14:[1, 14, 210, 3472,  65352, 1447768, 38853920, 1286816832, ...];
A^16:[1, 16, 272, 5024, 103456, 2441792, 67832064, 2265188480, ...]; ...
then the diagonal in the above table generates this sequence shift left:
[1/1, 4/2, 42/3, 784/4, 21760/5, 815184/6, 38853920/7, 2265188480/8, ...].
SUMS OF TERM RESIDUES MODULO 2^n.
Given a(k) == 0 (mod 2^n) for k>=(2*n-1) for n>=3, then it is interesting to consider the sums of the residues of all terms modulo 2^n for n>=1.
Let b(n) = Sum_{k>=0} a(k) (mod 2^n) for n>=1, then the sequence {b(n)} begins:
[2, 6, 14, 30, 62, 158, 350, 990, 2782, 5342, 10462, 16606, 61662, 119006, 233694, 495838, 1151198, 2592990, 5476574, 14389470, 30118110, 51089630, ...].

Links

Crossrefs

Cf. A233319, A231619, A231866, A231899, A232694, A232695, A232696.

Programs

  • PARI
    {a(n)=local(A=1+x); for(i=1, n, A=1+intformal(subst(A^2, x, x*A' +x*O(x^n)))); n!*polcoeff(A, n)}
for(n=0, 25, print1(a(n), ", "))
  • PARI
    {a(n)=local(A=1+x); for(i=1, n, A=1+intformal(1/x*serreverse(x/A^2 +x*O(x^n)))); n!*polcoeff(A, n)}
for(n=0, 25, print1(a(n), ", "))

Formula

E.g.f. A(x) satisfies: A(x)^2 = A'(x/A(x)^2).
E.g.f. A(x) satisfies: A(x) = sqrt( x / Series_Reversion( x*A'(x) ) ).
a(n) = [x^(n-1)/(n-1)!] A(x)^(2*n)/n for n>=1.

A233319 E.g.f. A(x) satisfies: A'(x) = A(x*A'(x))^3.

Original entry on oeis.org

1, 1, 3, 33, 726, 25236, 1229328, 78167484, 6193726506, 592068123054, 66673324219176, 8685890001564984, 1290531658541292252, 216188985806157611520, 40449991773179254230432, 8386998677130790903212000, 1914263814914709029067344724, 478208364783447353623777136772
Offset: 0

Views

Author

Paul D. Hanna, Dec 07 2013

Keywords

Comments

CONJECTURES:
a(n) == 0 (mod 2) for n>=4.
a(n) == 0 (mod 2^2) for n>=10.
a(n) == 0 (mod 2^3) for n>=18.
a(n) == 0 (mod 2^k) for n>=(8*k-6), k>=2.
a(n) == 0 (mod 3) for n>=2.
a(n) == 0 (mod 3^2) for n>=5.
a(n) == 0 (mod 3^3) for n>=7.
a(n) == 0 (mod 3^4) for n>=10.
a(n) == 0 (mod 3^k) for n>=(3*k-2), k>=3.
a(n) == 0 (mod 13) for n>=13.

Examples

			E.g.f.: A(x) = 1 + x + 3*x^2/2! + 33*x^3/3! + 726*x^4/4! + 25236*x^5/5! +...
such that
A(x*A'(x))^3 = A'(x) = 1 + 3*x + 33*x^2/2! + 726*x^3/3! + 25236*x^4/4! +...
A(x*A'(x)) = (A'(x))^(1/3) = 1 + x + 9*x^2/2! + 186*x^3/3! + 6330*x^4/4! + 306846*x^5/5! + 19560006*x^6/6! + 1559472498*x^7/7! +...
To illustrate a(n) = [x^(n-1)/(n-1)!] A(x)^(3*n)/n, create a table of coefficients of x^k/k!, k>=0, in A(x)^(3*n) like so:
A^3: [1,  3,  15,   159,   3240,   106218,   4961250,   305900982, ...];
A^6: [1,  6,  48,   588,  11646,   357336,  15709968,   923153004, ...];
A^9: [1,  9,  99,  1449,  30078,   899964,  37750104,  2118453588, ...];
A^12:[1, 12, 168,  2904,  65340,  1977912,  80833248,  4365682056, ...];
A^15:[1, 15, 255,  5115, 126180,  3961350, 161145630,  8476536330, ...];
A^18:[1, 18, 360,  8244, 223290,  7375968, 304020000, 15786282132, ...];
A^21:[1, 21, 483, 12453, 369306, 12932136, 547172388, 28405637064, ...];
A^24:[1, 24, 624, 17904, 578808, 21554064, 944463744, 49549812048, ...]; ...
then the diagonal in the above table generates this sequence shift left:
[1/1, 6/2, 99/3, 2904/4, 126180/5, 7375968/6, 547172388/7, 49549812048/8, ...].
SUMS OF TERM RESIDUES MODULO 2^n.
Given a(k) == 0 (mod 2^n) for k>=(8*n-6) for n>=2, then it is interesting to consider the sums of the residues of all terms modulo 2^n for n>=1.
Let b(n) = Sum_{k>=0} a(k) (mod 2^n) for n>=1, then the sequence {b(n)} begins:
[4, 12, 40, 112, 336, 848, 2128, 5584, 13776, 29648, 64464, 136144, 316368, ...].
SUMS OF TERM RESIDUES MODULO 3^n.
Given a(k) == 0 (mod 3^n) for k>=(3*n-2) for n>=3, then it is interesting to consider the sums of the residues of all terms modulo 3^n for n>=1.
Let c(n) = Sum_{k>=0} a(k) (mod 3^n) for n>=1, then the sequence {c(n)} begins:
[2, 17, 71, 368, 1340, 4985, 13733, 59660, 217124, 689516, 2520035, 6594416, 18286118, 72493100, 206416232, 722976884, 2617032608, 8170059617, 25603981622, 93015146708, 256894013555, 832213439720, 2338504300952, 6292517811686, 24650437682951, 71251311202316, 249181919185346, 729594560739527, ...].

Links

Crossrefs

Cf. A232686, A231619, A231866, A231899, A232694, A232695, A232696.

Programs

  • PARI
    {a(n)=local(A=1+x); for(i=1, n, A=1+intformal(subst(A^3, x, x*A' +x*O(x^n)))); n!*polcoeff(A, n)}
for(n=0, 25, print1(a(n), ", "))
  • PARI
    {a(n)=local(A=1+x); for(i=1, n, A=1+intformal(1/x*serreverse(x/A^3 +x*O(x^n)))); n!*polcoeff(A, n)}
for(n=0, 25, print1(a(n), ", "))

Formula

E.g.f. A(x) satisfies: A(x)^3 = A'(x/A(x)^3).
E.g.f. A(x) satisfies: A(x) = ( x / Series_Reversion( x*A'(x) ) )^(1/3).
a(n) = [x^(n-1)/(n-1)!] A(x)^(3*n)/n for n>=1.
Showing 1-6 of 6 results.

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