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.

A317346 O.g.f. A(x) satisfies: [x^n] exp( n^3*x - n^2*A(x) ) = 0 for n >= 1.

Original entry on oeis.org

1, 2, 72, 8096, 1839000, 695334816, 392764566208, 309340607492096, 323795915817507936, 434750954619876448000, 728547799352068864173632, 1490865523016798790557180928, 3659466509860384349989504297344, 10614823215131644149237135937187328, 35927108634064565449228268842108588800, 140351379904337650357154561973550135705600
Offset: 1

Views

Author

Paul D. Hanna, Jul 26 2018

Keywords

Comments

It is remarkable that this sequence should consist entirely of integers.

Examples

			O.g.f.: A(x) = x + 2*x^2 + 72*x^3 + 8096*x^4 + 1839000*x^5 + 695334816*x^6 + 392764566208*x^7 + 309340607492096*x^8 + ...
such that [x^n] exp( n^3*x - n^2*A(x) ) = 0  for n >= 1.
ILLUSTRATION OF DEFINITION.
The table of coefficients of x^k/k! in exp( n^3*x - n^2*A(x) ) begins:
n=1: [1, 0, -4, -432, -194256, -220662720, -500627544000, ...];
n=2: [1, 4, 0, -1856, -805376, -898258176, -2023715201024, ...];
n=3: [1, 18, 288, 0, -1989792, -2154563712, -4727980751616, ...];
n=4: [1, 48, 2240, 94464, 0, -4244861952, -9137589559296, ...];
n=5: [1, 100, 9900, 959200, 84852400, 0, -15901448888000, ...];
n=6: [1, 180, 32256, 5738688, 1003636224, 161358324480, 0, ...];
n=7: [1, 294, 86240, 25218144, 7335234144, 2103824749824, 557359956846336, 0, ...]; ...
in which the coefficient of x^n in row n forms a diagonal of zeros.
RELATED SERIES.
exp(A(x)) = 1 + x + 5*x^2/2! + 445*x^3/3! + 196105*x^4/4! + 221673401*x^5/5! + 501981700621*x^6/6! + 1983064113021685*x^7/7!  + ... + A317345(n)*x^n/n! + ...

Links

Crossrefs

Cf. A317344, A317345, A317347, A319941, A319942, A319943, A319944.

Programs

  • PARI
    {a(n) = my(A=[1], m); for(i=1, n+1, m=#A; A=concat(A, 0); A[m+1] = Vec( exp(m^3*x +x*O(x^#A)) / Ser(A)^(m^2) )[m+1]/m^2 ); polcoeff( log(Ser(A)),n)}
for(n=1,20,print1(a(n),", "))

Formula

a(n) ~ sqrt(1 - c) * 3^(3*n - 7/3) * n^(2*n - 5/2) / (sqrt(2*Pi) * exp(2*n) * c^(n - 1/3) * (3 - c)^(2*n - 2)), where c = -LambertW(-3*exp(-3)) = 0.1785606278779211065968086697... = -A226750. - Vaclav Kotesovec, Aug 07 2018

A317343 E.g.f. A(x) satisfies: [x^n] exp(n^2*x) / A(x)^n = 0 for n >= 1.

Original entry on oeis.org

1, 1, 3, 43, 1945, 178041, 26792971, 5940440563, 1812303908913, 725818277844145, 368664266359757971, 231291789356545214331, 175516846850044015048393, 158383499766971409675254953, 167568661592262324239839114395, 205391322843896611716680298647491, 288693442696565330249751284373878881, 461220054642000786943064376254671287393
Offset: 0

Views

Author

Paul D. Hanna, Jul 26 2018

Keywords

Comments

It is remarkable that the logarithm of the e.g.f. A(x) should be an integer series.

Examples

			E.g.f.: A(x) = 1 + x + 3*x^2/2! + 43*x^3/3! + 1945*x^4/4! + 178041*x^5/5! + 26792971*x^6/6! + 5940440563*x^7/7! + 1812303908913*x^8/8! + ...
ILLUSTRATION OF DEFINITION.
The table of coefficients of x^k/k! in exp(n^2*x) / A(x)^n begins:
n=1: [1, 0, -2, -36, -1764, -167280, -25620600, -5737974480, ...];
n=2: [1, 2, 0, -88, -4160, -371328, -55329536, -12201990400, ...];
n=3: [1, 6, 30, 0, -7812, -698184, -97733304, -20791334880, ...];
n=4: [1, 12, 136, 1296, 0, -1171968, -168658176, -33909447168, ...];
n=5: [1, 20, 390, 7220, 113020, 0, -265712600, -55963975600, ...];
n=6: [1, 30, 888, 25704, 709056, 16600320, 0, -84622337280, ...];
n=7: [1, 42, 1750, 72072, 2909340, 112245672, 3684715944, 0, ...];
n=8: [1, 56, 3120, 172640, 9455488, 508540416, 26199517696, 1150524892160, 0, ...]; ...
in which the coefficient of x^n in row n forms a diagonal of zeros.
LOGARITHMIC PROPERTY.
The logarithm of the e.g.f. is the integer series:
log(A(x)) = x + x^2 + 6*x^3 + 74*x^4 + 1400*x^5 + 35676*x^6 + 1140328*x^7 + 43740848*x^8 + 1954336608*x^9 + 99561874080*x^10 + ... + A317344(n)*x^n + ...

Links

Crossrefs

Cf. A317344, A317345.

Programs

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

Formula

a(n) ~ sqrt(1-c) * 2^(2*n - 3/2) * n^(2*n-1) / (exp(2*n) * c^(n - 1/2) * (2-c)^(n-1)), where c = -LambertW(-2*exp(-2)) = -A226775 = 0.4063757399599599... - Vaclav Kotesovec, Aug 06 2018

A319942 O.g.f. A(x) satisfies: [x^n] exp( n^4*x - n^2*A(x) ) = 0 for n >= 1.

Original entry on oeis.org

1, 18, 5616, 6776352, 20200266000, 119799079486272, 1242032508354758400, 20634722510624457007104, 515736410631216295520236032, 18480208067078637967802351884800, 913939154183946975187574927409795072, 60487962958244860971401604975128195088384
Offset: 1

Views

Author

Paul D. Hanna, Oct 02 2018

Keywords

Comments

It is remarkable that this sequence should consist entirely of integers.

Examples

			G.f.: A(x) = x + 18*x^2 + 5616*x^3 + 6776352*x^4 + 20200266000*x^5 + 119799079486272*x^6 + 1242032508354758400*x^7 + ...
ILLUSTRATION OF DEFINITION.
The table of coefficients of x^k/k! in exp( n^4*x - n^2*A(x) ) begins:
n=1: [1, 0, -36, -33696, -162628560, -2424019789440, ...];
n=2: [1, 12, 0, -138240, -657040896, -9735157974528, ...];
n=3: [1, 72, 4860, 0, -1533920976, -22357116073728, ...];
n=4: [1, 240, 57024, 12870144, 0, -41496660080640, ...];
n=5: [1, 600, 359100, 213537600, 121570858800, 0, ...];
n=6: [1, 1260, 1586304, 1994264064, 2496165050880, 3006510865205760, 0, ...]; ...
in which the coefficient of x^n in row n forms a diagonal of zeros.
RELATED SERIES.
exp(A(x)) = 1 + x + 37*x^2/2! + 33805*x^3/3! + 162771337*x^4/4! + 2424857569561*x^5/5! + 86269983111064621*x^6/6! + 6260449705448367386917*x^7/7! + ...

Links

Crossrefs

Cf. A319941, A319943, A319944, A317344, A317345, A317346, A317347.

Programs

  • PARI
    {a(n) = my(A=[1], m); for(i=1, n+1, m=#A; A=concat(A, 0); A[m+1] = Vec( exp(m^4*x +x*O(x^#A)) / Ser(A)^(m^2) )[m+1]/m^2 ); polcoeff( log(Ser(A)), n)}
for(n=1, 15, print1(a(n), ", "))

Formula

a(n) ~ sqrt(1-c) * 2^(8*n - 9/2) * n^(3*n - 5/2) / (sqrt(Pi) * c^n * (4-c)^(3*n - 2) * exp(3*n)), where c = -LambertW(-4*exp(-4)) = 0.07930960512711365643910864738... - Vaclav Kotesovec, Oct 13 2020
Showing 1-3 of 3 results.

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