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 1-5 of 5 results.

A296173 G.f. equals the logarithm of the e.g.f. of A296172.

Original entry on oeis.org

1, -3, -30, -2686, -517311, -173118807, -88535206152, -63977172334344, -61971659588102940, -77470793599569049440, -121439997599825393413344, -233353875172602479932391040, -539638027429765922735002220880, -1479049138515818646669055218090480, -4742815067612592169849894663392228480, -17597031102801426396121130730318359114880, -74817150772352720408567833273371047298417408
Offset: 1

Views

Author

Paul D. Hanna, Dec 07 2017

Keywords

Comments

E.g.f. G(x) of A296172 satisfies: [x^(n-1)] G(x)^(n^3) = [x^n] G(x)^(n^3) for n>=1.

Examples

			G.f. A(x) = x - 3*x^2 - 30*x^3 - 2686*x^4 - 517311*x^5 - 173118807*x^6 - 88535206152*x^7 - 63977172334344*x^8 - 61971659588102940*x^9 - 77470793599569049440*x^10 - 121439997599825393413344*x^11 - 233353875172602479932391040*x^12 - 539638027429765922735002220880*x^13 - 1479049138515818646669055218090480*x^14 - 4742815067612592169849894663392228480*x^15 +...
such that
G(x) = exp(A(x)) = 1 + x - 5*x^2/2! - 197*x^3/3! - 65111*x^4/4! - 62390159*x^5/5! - 125012786669*x^6/6! - 447082993406405*x^7/7! - 2583111044504384687*x^8/8! - 22511408975342644804991*x^9/9! - 281350305428215911326408789*x^10/10! - 4850582201056517165575319399909*x^11/11! - 111834955668396093904661955538037255*x^12/12! +...
satisfies [x^(n-1)] G(x)^(n^3) = [x^n] G(x)^(n^3) for n>=1.
Series_Reversion(A(x)) = x + 3*x^2 + 48*x^3 + 3271*x^4 + 575163*x^5 + 185377116*x^6 + 93039467356*x^7 + 66505075585875*x^8 + 63970743282062646*x^9 + 79580632411431634441*x^10 + 124299284968805234137968*x^11 + 238188439678208173206500760*x^12 +...+ A295813(n)*x^n +...
		

Crossrefs

Programs

  • PARI
    {a(n) = my(A=[1]); for(i=1,n+1, A=concat(A,0); V=Vec(Ser(A)^((#A-1)^3)); A[#A] = (V[#A-1] - V[#A])/(#A-1)^3 ); polcoeff(log(Ser(A)),n)}
    for(n=1,30,print1(a(n),", "))

Formula

a(n) ~ -sqrt(1-c) * 3^(3*n - 3) * n^(2*n - 7/2) / (sqrt(2*Pi) * c^n * (3-c)^(2*n - 3) * exp(2*n)), where c = -LambertW(-3*exp(-3)) = -A226750. - Vaclav Kotesovec, Oct 13 2020

A295813 G.f. A(x) satisfies: G(A(x)) = exp(x), where G(x) equals the e.g.f. of A296172.

Original entry on oeis.org

1, 3, 48, 3271, 575163, 185377116, 93039467356, 66505075585875, 63970743282062646, 79580632411431634441, 124299284968805234137968, 238188439678208173206500760, 549611050835556942751087049225, 1503700734638162443238902233252144, 4814751647416985610768723994195186728, 17841762828286483988438913318683740082187, 75777421917902616009655480827109144353730842
Offset: 1

Views

Author

Paul D. Hanna, Dec 09 2017

Keywords

Comments

E.g.f. G(x) of A296172 satisfies: [x^(n-1)] G(x)^(n^3) = [x^n] G(x)^(n^3) for n>=1.

Examples

			G.f.: A(x) = x + 3*x^2 + 48*x^3 + 3271*x^4 + 575163*x^5 + 185377116*x^6 + 93039467356*x^7 + 66505075585875*x^8 + 63970743282062646*x^9 + 79580632411431634441*x^10 + 124299284968805234137968*x^11 + 238188439678208173206500760*x^12 +...
The series reversion equals the logarithm of the e.g.f. of A296172, which begins:
Series_Reversion(A(x)) = x - 3*x^2 - 30*x^3 - 2686*x^4 - 517311*x^5 - 173118807*x^6 - 88535206152*x^7 - 63977172334344*x^8 - 61971659588102940*x^9 - 77470793599569049440*x^10 - 121439997599825393413344*x^11 - 233353875172602479932391040*x^12 +...+ A296173(n)*x^n +...
		

Crossrefs

Programs

  • PARI
    {a(n) = my(A=[1]); for(i=1, n+1, A=concat(A, 0); V=Vec(Ser(A)^((#A-1)^3)); A[#A] = (V[#A-1] - V[#A])/(#A-1)^3 ); polcoeff(serreverse(log(Ser(A))), n)}
    for(n=1, 30, print1(a(n), ", "))

Formula

G.f. is the series reversion of the logarithm of the e.g.f. of A296172.
a(n) ~ sqrt(1-c) * 3^(3*n - 3) * n^(2*n - 7/2) / (sqrt(2*Pi) * c^n * (3-c)^(2*n - 3) * exp(2*n)), where c = -LambertW(-3*exp(-3)) = -A226750. - Vaclav Kotesovec, Oct 13 2020

A296170 E.g.f. A(x) satisfies: [x^(n-1)] A(x)^(n^2) = [x^n] A(x)^(n^2) for n>=1.

Original entry on oeis.org

1, 1, -1, -11, -239, -17059, -2145689, -412595231, -111962826751, -40590007936199, -18900753214178609, -10974885891916507219, -7765167486697279401071, -6571694718107813687003051, -6551841491106355785902247049, -7597507878436131044487467850599, -10136619271768255373949409579309439, -15416099624633773180711565727641136271, -26508391106594400233543066679525341764961
Offset: 0

Views

Author

Paul D. Hanna, Dec 07 2017

Keywords

Comments

Compare e.g.f. to: [x^(n-1)] exp(x)^n = [x^n] exp(x)^n for n>=1.

Examples

			E.g.f.: A(x) = 1 + x - x^2/2! - 11*x^3/3! - 239*x^4/4! - 17059*x^5/5! - 2145689*x^6/6! - 412595231*x^7/7! - 111962826751*x^8/8! - 40590007936199*x^9/9! - 18900753214178609*x^10/10! - 10974885891916507219*x^11/11! - 7765167486697279401071*x^12/12! - 6571694718107813687003051*x^13/13! - 6551841491106355785902247049*x^14/14! - 7597507878436131044487467850599*x^15/15! +...
To illustrate [x^(n-1)] A(x)^(n^2) = [x^n] A(x)^(n^2), form a table of coefficients of x^k in A(x)^(n^2) that begins as
n=1: [(1), (1), -1/2, -11/6, -239/24, -17059/120, -2145689/720, ...];
n=2: [1, (4), (4), -28/3, -196/3, -10472/15, -614264/45, ...];
n=3: [1, 9, (63/2), (63/2), -1701/8, -98217/40, -3168081/80, ...];
n=4: [1, 16, 112, (1232/3), (1232/3), -95648/15, -4835264/45, ...];
n=5: [1, 25, 575/2, 11725/6, (190225/24), (190225/24), ...];
n=6: [1, 36, 612, 6444, 45684, (1043784/5), (1043784/5), ...];
n=7: [1, 49, 2303/2, 102949/6, 4313617/24, 164086349/120, (5086480231/720), (5086480231/720), ...];
...
in which the diagonals indicated by parenthesis are equal.
Dividing the coefficients of x^(n-1)/(n-1)! in A(x)^(n^2) by n^2, we obtain the following sequence:
[1, 1, 7, 154, 7609, 695856, 103805719, 23134327168, 7227250033329, 3017857024161280, 1623903877812828871, ..., A296232(n), ...].
LOGARITHMIC PROPERTY.
Amazingly, the logarithm of the e.g.f. A(x) is an integer series:
log(A(x)) = x - x^2 - x^3 - 9*x^4 - 134*x^5 - 2852*x^6 - 79096*x^7 - 2699480*x^8 - 109201844*x^9 - 5100872244*x^10 - 269903909820*x^11 - 15944040740604*x^12 - 1039553309158964*x^13 - 74123498185170292*x^14 - 5736368141560365292*x^15 - 478780244956262592748*x^16 - 42865943103053965559668*x^17 - 4097785410628237071311764*x^18 - 416572537937169684523985420*x^19 - 44873737158384968851319470220*x^20 +...+ A296171(n)*x^n +...
		

Crossrefs

Programs

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

Formula

The logarithm of the e.g.f. A(x) is an integer series:
log(A(x)) = Sum{n>=1} A296171(n) * x^n.
E.g.f. A(x) satisfies:
_ 1/n! * d^n/dx^n A(x)^(n^2) = 1/(n-1)! * d^(n-1)/dx^(n-1) A(x)^(n^2) for n>=1, when evaluated at x = 0.
a(n) ~ c * d^n * n^(2*n-2) / exp(2*n), where d = -4/(LambertW(-2*exp(-2))*(2+LambertW(-2*exp(-2)))) = 6.17655460948348035823168... and c = -0.1875440087... - Vaclav Kotesovec, Dec 23 2017

A296174 E.g.f. A(x) satisfies: [x^(n-1)] A(x)^(n^4) = [x^n] A(x)^(n^4) for n>=1.

Original entry on oeis.org

1, 1, -13, -2999, -8197751, -81738176899, -2105524335759389, -115916378979693710123, -12069952631345502122877199, -2179911119857340269414590758951, -639738016495616440994202167765715629, -289812262583683385183617291938537580840159, -194420626455357631368336026954933981532680935943, -186615832949734453391125561079799823405868770406129579
Offset: 0

Views

Author

Paul D. Hanna, Dec 07 2017

Keywords

Comments

Compare e.g.f. to: [x^(n-1)] exp(x)^n = [x^n] exp(x)^n for n>=1.

Examples

			E.g.f.: A(x) = 1 + x - 13*x^2/2! - 2999*x^3/3! - 8197751*x^4/4! - 81738176899*x^5/5! - 2105524335759389*x^6/6! - 115916378979693710123*x^7/7! - 12069952631345502122877199*x^8/8! - 2179911119857340269414590758951*x^9/9! - 639738016495616440994202167765715629*x^10/10! +...
To illustrate [x^(n-1)] A(x)^(n^4) = [x^n] A(x)^(n^4), form a table of coefficients of x^k in A(x)^(n^4) that begins as
n=1: [(1), (1), -13/2, -2999/6, -8197751/24, -81738176899/120, ...];
n=2: [1, (16), (16), -26992/3, -16767472/3, -164706495728/15, ...];
n=3: [1, 81, (5427/2), (5427/2), -246155517/8, -2300014714833/40, ...];
n=4: [1, 256, 30976, (6633728/3), (6633728/3), -2939838787328/15, ...];
n=5: [1, 625, 381875/2, 225885625/6, (122571375625/24), (122571375625/24), ...];
n=6: [1, 1296, 830736, 350400816, 108698540976, (126219948303024/5), (126219948303024/5), ...];
...
in which the diagonals indicated by parenthesis are equal.
Dividing the coefficients of x^(n-1)/(n-1)! in A(x)^(n^4) by n^4, we obtain the following sequence:
[1, 1, 67, 51826, 196114201, 2337406450056, 68145136372652611, 4136219111307043556272, 467591060765602023501093201, ...].
LOGARITHMIC PROPERTY.
Amazingly, the logarithm of the e.g.f. A(x) is an integer series:
log(A(x)) = x - 7*x^2 - 493*x^3 - 341101*x^4 - 680813601*x^5 - 2923660883625*x^6 - 22996362478599551*x^7 - 299331006952284448127*x^8 - 6006951481145880962408552*x^9 - 176288642409787912257773903552*x^10 - 7260231964238768891891716773249396*x^11 - 405879958110794676900559524931590299892*x^12 +...
		

Crossrefs

Programs

  • PARI
    {a(n) = my(A=[1]); for(i=1,n+1, A=concat(A,0); V=Vec(Ser(A)^((#A-1)^4)); A[#A] = (V[#A-1] - V[#A])/(#A-1)^4 ); n!*A[n+1]}
    for(n=0,30,print1(a(n),", "))

Formula

The logarithm of the e.g.f. A(x) is an integer series:
log(A(x)) = Sum{n>=1} A296175(n) * x^n.
E.g.f. A(x) satisfies:
_ 1/n! * d^n/dx^n A(x)^(n^4) = 1/(n-1)! * d^(n-1)/dx^(n-1) A(x)^(n^4) for n>=1, when evaluated at x = 0.
a(n) ~ -sqrt(1-c) * 2^(8*n - 8) * n^(4*n - 4) / (c^n * (4-c)^(3*n - 4) * exp(4*n)), where c = -LambertW(-4*exp(-4)) = 0.079309605127113656439108647386463779474372... - Vaclav Kotesovec, Oct 13 2020

A296176 E.g.f. A(x) satisfies: [x^(n-1)] A(x)^(n^5) = [x^n] A(x)^(n^5) for n>=1.

Original entry on oeis.org

1, 1, -29, -36629, -734559239, -71200423546199, -22459270436075644469, -18407129959728493123679069, -33747438879000326056232288023439, -124162549312926509293620790889452447919, -843670934957017748849439817665935283173590349, -9914324850699841477684471316247032518786477385700389, -191047752973105011101288266443568575709649708408401069796759
Offset: 0

Views

Author

Paul D. Hanna, Dec 07 2017

Keywords

Comments

Compare e.g.f. to: [x^(n-1)] exp(x)^n = [x^n] exp(x)^n for n>=1.

Examples

			E.g.f.: A(x) = 1 + x - 29*x^2/2! - 36629*x^3/3! - 734559239*x^4/4! - 71200423546199*x^5/5! - 22459270436075644469*x^6/6! - 18407129959728493123679069*x^7/7! - 33747438879000326056232288023439*x^8/8! - 124162549312926509293620790889452447919*x^9/9! - 843670934957017748849439817665935283173590349*x^10/10! +...
To illustrate [x^(n-1)] A(x)^(n^5) = [x^n] A(x)^(n^5), form a table of coefficients of x^k in A(x)^(n^5) that begins as
n=1: [(1), (1), -29/2, -36629/6, -734559239/24, -71200423546199/120, ...];
n=2: [1, (32), (32), -614336/3, -2956631488/3, -285257147669696/15, ...];
n=3: [1, 243, (51759/2), (51759/2), -62010059733/8, -5840748850240719/40, ...];
n=4: [1, 1024, 508928, (470976512/3), (470976512/3), -9540780758505472/15, ...];
n=5: [1, 3125, 9671875/2, 29524484375/6, (86178242265625/24), (86178242265625/24), ...];
n=6: [1, 7776, 30116448, 77409815616, 148214160396864, (1099707612312815424/5), (1099707612312815424/5), ...];
...
in which the diagonals indicated by parenthesis are equal.
Dividing the coefficients of x^(n-1)/(n-1)! in A(x)^(n^5) by n^5, we obtain the following sequence:
[1, 1, 213, 919876, 27577037525, 3394159297261776, 1269158820664910885737, 1186717596374463676630699264, ...].
LOGARITHMIC PROPERTY.
Amazingly, the logarithm of the e.g.f. A(x) is an integer series:
log(A(x)) = x - 15*x^2 - 6090*x^3 - 30600650*x^4 - 593306350650*x^5 - 31192838317208826*x^6 - 3652177141294409632400*x^7 - 836986399841753367052602000*x^8 - 342157863774785896821739864893375*x^9 - 232492750600387706453977026534258393375*x^10 +...
		

Crossrefs

Programs

  • PARI
    {a(n) = my(A=[1]); for(i=1,n+1, A=concat(A,0); V=Vec(Ser(A)^((#A-1)^5)); A[#A] = (V[#A-1] - V[#A])/(#A-1)^5 ); n!*A[n+1]}
    for(n=0,30,print1(a(n),", "))

Formula

The logarithm of the e.g.f. A(x) is an integer series:
log(A(x)) = Sum{n>=1} A296177(n) * x^n.
E.g.f. A(x) satisfies:
_ 1/n! * d^n/dx^n A(x)^(n^5) = 1/(n-1)! * d^(n-1)/dx^(n-1) A(x)^(n^5) for n>=1, when evaluated at x = 0.
Showing 1-5 of 5 results.