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 1-9 of 9 results.

A341871 Coefficients of the series whose 48th power equals E_2(x)^2/E_4(x), where E_2(x) and E_4(x) are the Eisenstein series A006352 and A004009.

Original entry on oeis.org

1, -6, 558, -88884, 15433662, -2864048616, 552921962724, -109731286565040, 22220439670517814, -4569456313225317114, 951159953810624453208, -199945837161334089352548, 42373766861587365894611604
Offset: 0

Views

Author

Peter Bala, Feb 22 2021

Keywords

Comments

It is easy to see that E_2(x)^2/E_4(x) == 1 - 48*Sum_{k >= 1} (k + 5*k^3)*x^k/(1 - x^k) (mod 288), and also that the integer k + 5*k^3 is always divisible by 6. Hence, E_2(x)^2/E_4(x) == 1 (mod 288). It follows from Heninger et al., p. 3, Corollary 2, that the series expansion of (E_2(x)^2/E_4(x))^(1/48) = 1 - 6*x + 558*x^2 - 88884*x^3 + 15433662*x^4 - ... has integer coefficients.
Note that (E_2(x)^2/E_4(x))^(1/48) = (E_2(x)^4/E_8(x))^(1/96).

Crossrefs

Programs

  • Maple
    E(2,x) := 1 -  24*add(k*x^k/(1-x^k),   k = 1..20):
    E(4,x) := 1 + 240*add(k^3*x^k/(1-x^k), k = 1..20):
    with(gfun): series((E(2,x)^2/E(4,x))^(1/48), x, 20):
    seriestolist(%);

A377973 Expansion of the 96th root of the series 2*E_2(x) - E_2(x)^2, where E_2 is the Eisenstein series of weight 2.

Original entry on oeis.org

1, 0, -6, -36, -1812, -20748, -773340, -12237456, -386587650, -7368446268, -211914644940, -4517757977820, -123221458979940, -2814502962357420, -74551748141034552, -1778129476480366320, -46377354051910716180, -1137191336376638407704, -29438532048777299115090, -735051729258136807204140
Offset: 0

Views

Author

Peter Bala, Nov 13 2024

Keywords

Comments

Let R = 1 + x*Z[[x]] denote the set of integer power series with constant term equal to 1. Let P(n) = {g^n, g in R}. The Eisenstein series E_2(x) lies in P(4) (Heninger et al.). Hence E_2(x)^2 lies in P(8).
We claim that the series 2*E_2(x) - E_2(x)^2 belongs to P(96).
Proof.
E_2(x) = 1 - 24*Sum_{n >= 1} sigma_1(n)*x^n.
Hence,
2*E_2(x) - E_2(x)^2 = 1 - (24^2)*(Sum_{n >= 1} sigma_1(n)*x^n )^2 is in the set R.
Hence, 2*E_2(x) - E_2(x)^2 == 1 (mod 24^2) == 1 (mod (2^6)*(3^2)).
It follows from Heninger et al., Theorem 1, Corollary 2, that the series 2*E_2(x) - E_2(x)^2 belongs to P((2^5)*3) = P(96). End Proof.

Crossrefs

Cf. A006352 (E_2), A281374 (E_2)^2, A289392 ((E_2)^(1/4)), A341801, A341871 - A341875, A377974, A377975, A377976, A377977.

Programs

  • Maple
    with(numtheory):
    E := proc (k) local n, t1; t1 := 1 - 2*k*add(sigma[k-1](n)*q^n, n = 1..30)/bernoulli(k); series(t1, q, 30) end:
    seq(coeftayl((2*E(2) - E(2)^2)^(1/96), q = 0, n),n = 0..20);
  • Mathematica
    terms = 20; E2[x_] = 1 - 24*Sum[k*x^k/(1 - x^k), {k, 1, terms}]; CoefficientList[Series[(2*E2[x] - E2[x]^2)^(1/96), {x, 0, terms}], x] (* Vaclav Kotesovec, Aug 03 2025 *)

Formula

a(n) ~ c / (r^n * n^(97/96)), where r = A211342 = 0.03727681029645165815098... and c = -0.0104397599261506010365791466642760245638473040812140699981294533624... - Vaclav Kotesovec, Aug 03 2025

A377974 Expansion of the 1920th root of the series 2*E_4(x) - E_8(x), where E_4 and E_8 are the Eisenstein series of weight 4 and weight 8.

Original entry on oeis.org

1, 0, -30, -540, -867660, -31107300, -33668157900, -1795572812400, -1477793386682970, -103845834995498100, -69550699526934273180, -6017200267937951322660, -3426636160378174348594500, -349303370036461528632524580, -174458882971934188146144343320, -20314204536496741742949242177040
Offset: 0

Views

Author

Peter Bala, Nov 13 2024

Keywords

Comments

Let R = 1 + x*Z[[x]] denote the set of integer power series with constant term equal to 1. Let P(n) = {g^n, g in R}. The Eisenstein series E_4(x) lies in P(8) (Heninger et al.). Since E_8(x) = E_4(x)^2, it follows that E_8(x) lies in P(16).
We claim that the series 2*E_4(x) - E_8(x) belongs to P(1920).
Proof.
E_4(x) = 1 + 240*Sum_{n >= 1} sigma_3(n)*x^n. Hence,
2*E_4(x) - E_8(x) = 2*E_4(x) - E_4(x)^2 = 1 - 240^2*( Sum_{n >= 1} sigma_3(n) )^2 is in the set R.
Hence, 2*E_4(x) - E_8(x) == 1 mod(240^2) == 1 (mod (2^8)*(3^2)*(5^2)).
It follows from Heninger et al., Theorem 1, Corollary 2, that the series 2*E_4(x) - E_8(x) belongs to P((2^7)*3*5) = P(1920). End Proof.

Crossrefs

Cf. A004009 (E_4), A008410 (E_8), A108091 (eighth root of E_4), A341871 - A341875, A377973, A377975, A377976, A377977.

Programs

  • Maple
    with(numtheory):
    E := proc (k) local n, t1; t1 := 1 - 2*k*add(sigma[k-1](n)*q^n, n = 1..30)/bernoulli(k); series(t1, q, 30) end:
    seq(coeftayl((2*E(4) - E(8))^(1/1920), q = 0, n),n = 0..20);
  • Mathematica
    terms = 20; E4[x_] = 1 + 240*Sum[k^3*x^k/(1 - x^k), {k, 1, terms}]; E8[x_] = 1 + 480*Sum[k^7*x^k/(1 - x^k), {k, 1, terms}]; CoefficientList[Series[(2*E4[x] - E8[x])^(1/1920), {x, 0, terms}], x] (* Vaclav Kotesovec, Aug 03 2025 *)

Formula

a(n) ~ c / (r^n * n^(1921/1920)), where r = 0.004019427095115250686492968205049012182922598389629390919504184161606551652... is the root of the equation Sum_{k>=1} sigma_3(k) * r^k = 1/240 and c = -0.00052087420429807426289253718287... - Vaclav Kotesovec, Aug 03 2025

A377975 Expansion of the 6048th root of the series 2*E_6(x) - E_6(x)^2, where E_6 is the Eisenstein series of weight 6.

Original entry on oeis.org

1, 0, -42, -2772, -5399688, -704781084, -943173698460, -180121119486672, -188146584694894350, -46293152603021155692, -40574254265781269371884, -11963000065787771567311500, -9221266403646163252100062068, -3107813621461888912485774582588, -2176998806586925223600540321844120
Offset: 0

Views

Author

Peter Bala, Nov 14 2024

Keywords

Comments

Let R = 1 + x*Z[[x]] denote the set of integer power series with constant term equal to 1. Let P(n) = {g^n, g in R}. The Eisenstein series E_6(x) lies in P(12) (Heninger et al.).
We claim that the series 2*E_6(x) - E_6(x)^2 belongs to P(6048).
Proof.
E_6(x) = 1 - 504*Sum_{n >= 1} sigma_5(n)*x^n. Hence,
2*E_6(x) - E_6(x)^2 = 1 - (504^2)*( Sum_{n >= 1} sigma_5(n)*x^n )^2 is in R.
Hence, 2*E_6(x) - E_6(x)^2 == 1 (mod 504^2) == 1 (mod (2^6)*(3^4)*(7^2)).
It follows from Heninger et al., Theorem 1, Corollary 2, that the series 2*E_6(x) - E_6(x)^2 belongs to P((2^5)*(3^3)*7) = P(6048). End Proof.

Crossrefs

Cf. A013973 (E_6), A109817 ( (E_6)^1/12 ), A280869 (E_6)^2, A341871 - A341875, A377973, A377974, A377976, A377977.

Programs

  • Maple
    with(numtheory):
    E := proc (k) local n, t1; t1 := 1 - 2*k*add(sigma[k-1](n)*q^n, n = 1..30)/bernoulli(k); series(t1, q, 30) end:
    seq(coeftayl((2*E(6) - E(6)^2)^(1/6048), q = 0, n),n = 0..20);
  • Mathematica
    terms = 20; E6[x_] = 1 - 504*Sum[k^5*x^k/(1 - x^k), {k, 1, terms}]; CoefficientList[Series[(2*E6[x] - E6[x]^2)^(1/6048), {x, 0, terms}], x] (* Vaclav Kotesovec, Aug 03 2025 *)

Formula

a(n) ~ c / (r^n * n^(6049/6048)), where r = 0.0018674427317079888144302129348270303934228050024753171993815386383179351229... is the root of the equation Sum_{k>=1} sigma_5(k) * r^k = 1/504 and c = -0.0001653486643613776568861731992670297686378824546... - Vaclav Kotesovec, Aug 03 2025

A377977 Expansion of the 288th root of the series 3*E_4(x) - 2*E_6(x), where E_4(x) and E_6(x) are the Eisenstein series of weight 4 and 6.

Original entry on oeis.org

1, 6, -5028, 5704188, -7284893010, 9926715853068, -14092613175928308, 20580782244716567592, -30684764269418402550900, 46478269075227117026711730, -71284154421570122590465786956, 110437754516732491586466670733772, -172528135408494997625486967978486588, 271418933884659782820559630827037837908
Offset: 0

Views

Author

Peter Bala, Nov 15 2024

Keywords

Comments

Let R = 1 + x*Z[[x]] denote the set of integer power series with constant term equal to 1. Let P(n) = {g^n, g in R}. The Eisenstein series E_4(x) lies in P(8) and E_6(x) lies in P(12) (Heninger et al.).
We claim that the series 3*E_4(x) - 2*E_6(x) belongs to P(288).
Proof.
E_4(x) = 1 + 240*Sum_{n >= 1} sigma_3(n)*x^n.
E_6(x) = 1 - 504*Sum_{n >= 1} sigma_5(n)*x^n.
Hence, 3*E_4(x) - 2*E_6(x) = 1 - (12^3)*Sum_{n >= 1} (1/12)*(5*sigma_3(n) + 7*sigma_5(n))*x^n belong to R, since the polynomial (1/12)*(5*k^3 + 7*k^5) is integral for integer values of k. See A245380.
Hence, 3*E_4(x) - 2*E_6(x) == 1 (mod 12^3) == 1 (mod (2^6)*(3^3)).
It follows from Heninger et al., Theorem 1, Corollary 2, that the series 3*E_4(x) - 2*E_6(x) belongs to P((2^5)*(3^2)) = P(288). End Proof.

Crossrefs

Cf. A004009 (E_4), A013973 (E_6), A108091 ((E_4)^1/8), A109817 ((E_6)^1/12), A245380, A341871 - A341875, A377973, A377974, A377975.

Programs

  • Maple
    with(numtheory):
    E := proc (k) local n, t1; t1 := 1 - 2*k*add(sigma[k-1](n)*q^n, n = 1..30)/bernoulli(k); series(t1, q, 30) end:
    seq(coeftayl((3*E(4) - 2*E(6))^(1/288), q = 0, n), n = 0..20);
  • Mathematica
    terms = 20; E4[x_] = 1 + 240*Sum[k^3*x^k/(1 - x^k), {k, 1, terms}]; E6[x_] = 1 - 504*Sum[k^5*x^k/(1 - x^k), {k, 1, terms}]; CoefficientList[Series[(3*E4[x] - 2*E6[x])^(1/288), {x, 0, terms}], x] (* Vaclav Kotesovec, Aug 03 2025 *)

Formula

a(n) ~ (-1)^(n+1) * c * d^n / n^(289/288), where d = 1704.7780406875645261102091212390097973945883014209828800432529862899259963... and c = 0.0034650713031853295969588514070741337333119867976967661391075146399616... - Vaclav Kotesovec, Aug 03 2025

A377976 Expansion of the 48th root of the series 2*E_2(x) - E_4(x), where E_2(x) and E_4(x) are the Eisenstein series of weight 2 and 4.

Original entry on oeis.org

1, -6, -894, -174420, -38431614, -9048710040, -2221653118116, -561444889080960, -144914324838755910, -38011797621225586602, -10098281618881696696392, -2710458654395655881518356, -733711171629600485187568404, -200033609249999737396399900920, -54867682197669353983111639906656
Offset: 0

Views

Author

Peter Bala, Nov 14 2024

Keywords

Comments

Let R = 1 + x*Z[[x]] denote the set of integer power series with constant term equal to 1. Let P(n) = {g^n, g in R}. The Eisenstein series E_2(x) lies in P(4) and E_4(x) lies in P(8) (Heninger et al.).
We claim that the series 2*E_2(x) - E_4(x) belongs to P(48).
Proof.
E_2(x) = 1 - 24*Sum_{n >= 1} sigma_1(n)*x^n.
E_4(x) = 1 + 240*Sum_{n >= 1} sigma_3(n)*x^n.
Hence,
2*E_2(x) - E_4(x) = 1 - (288)*Sum_{n >= 1} ((1/6)*sigma_1(n) + (5/6)*sigma_3(n))*x^n belongs to the set R, since the polynomial (1/6)*k + (5/6)*k^3 has integer values for integer k. See A004068.
Hence, 2*E_2(x) - E_4(x) == 1 (mod 288) == 1 (mod (2^5)*(3^2)).
It follows from Heninger et al., Theorem 1, Corollary 2, that the series 2*E_2(x) - E_4(x) belongs to P((2^4)*3) = P(48). End Proof.
In a similar way we find that the series 3*E_2(x) - E_6(x) - 1 belongs to P(72) and the three series 3*E_4(x) - 2*E_6(x), 5*E_4(x) - 2*E_10(x) - 2 and 5*E_6(x) - 3*E_10(x) - 1 belong to P(288).

Crossrefs

Cf. A004068, A006352 (E_2), A004009 (E_4), A108091 ((E_4)^1/8), A289392 ((E_2)^(1/4)), A341871 - A341875, A377973, A377974, A377975, A377977.

Programs

  • Maple
    with(numtheory):
    E := proc (k) local n, t1; t1 := 1 - 2*k*add(sigma[k-1](n)*q^n, n = 1..30)/bernoulli(k); series(t1, q, 30) end:
    seq(coeftayl((2*E(2) - E(4))^(1/48), q = 0, n),n = 0..20);
  • Mathematica
    terms = 20; E2[x_] = 1 - 24*Sum[k*x^k/(1 - x^k), {k, 1, terms}]; E4[x_] = 1 + 240*Sum[k^3*x^k/(1 - x^k), {k, 1, terms}]; CoefficientList[Series[(2*E2[x] - E4[x])^(1/48), {x, 0, terms}], x] (* Vaclav Kotesovec, Aug 03 2025 *)

Formula

a(n) ~ c * d^n / n^(49/48), where d = 295.8669385406700495308385233671383399895922733900742171390678012914822364544611... and c = -0.0205882497833853345146399243734199945444083043388859856935627869352251231763... - Vaclav Kotesovec, Aug 03 2025

A341874 Coefficients of the series whose 24th power equals E_2(x)^7/E_14(x), where E_2(x) and E_14(x) are the Eisenstein series A006352 and A058550.

Original entry on oeis.org

1, -6, 8118, 1740636, 937783902, 364856395608, 172736345164500, 78278100914583312, 37268001893898954198, 17773741638825114790854, 8624927270409695050736952, 4214914849580580859932456300, 2078204723099375850950863499028
Offset: 0

Views

Author

Peter Bala, Feb 23 2021

Keywords

Comments

It is easy to see that E_2(x)^7/E_14(x) == 1 - 24*Sum_{k >= 1} (7*k - 11*k^13)*x^k/(1 - x^k) (mod 144), and also that the integer 7*k - k^13 is always divisible by 6. Hence, E_2(x)^7/E_14(x) == 1 (mod 144). It follows from Heninger et al., p. 3, Corollary 2, that the series expansion of (E_2(x)^7/E_14(x))^(1/24) = 1 - 6*x + 8118*x^2 + 1740636*x^3 + 937783902*x^4 + ... has integer coefficients.

Crossrefs

Programs

  • Maple
    E(2,x)  := 1 -  24*add(k*x^k/(1-x^k),   k = 1..20):
    E(14,x) := 1 - 24*add(k^13*x^k/(1-x^k), k = 1..20):
    with(gfun): series((E(2,x)^7/E(14,x))^(1/24), x, 20):
    seriestolist(%);

A341872 Coefficients of the series whose 72nd power equals E_2(x)^3/E_6(x), where E_2(x) and E_6(x) are the Eisenstein series A006352 and A013973.

Original entry on oeis.org

1, 6, 1998, 722484, 291762942, 125454173544, 56146411655460, 25832836404319152, 12128921727745915062, 5783583949613172902394, 2791762868052719757442008, 1360988846025232489401029220, 668925190887642335984231235348, 331039288912491308442251418152952
Offset: 0

Views

Author

Peter Bala, Feb 22 2021

Keywords

Comments

It is easy to see that E_2(x)^3/E_6(x) == 1 - 72*Sum_{k >= 1} (k - 7*k^5)*x^k/(1 - x^k) (mod 432), and also that the integer k - 7*k^5 is always divisible by 6. Hence, E_2(x)^3/E_6(x) == 1 (mod 432). It follows from Heninger et al., p. 3, Corollary 2, that the series expansion of (E_2(x)^2/E_6(x))^(1/72) = 1 + 6*x + 1998*x^2 + 722484*x^3 + 291762942* x^4 + ... has integer coefficients.

Crossrefs

Programs

  • Maple
    E(2,x) := 1 -  24*add(k*x^k/(1-x^k),   k = 1..20):
    E(6,x) := 1 - 504*add(k^5*x^k/(1-x^k), k = 1..20):
    with(gfun): series((E(2,x)^3/E(6,x))^(1/72), x, 20):
    seriestolist(%);

Formula

a(n) ~ c * exp(2*Pi*n) / n^(71/72), where c = 0.013960369132490470055158573616810629626490780934389076244815126342923645628... - Vaclav Kotesovec, Mar 08 2021

A341873 Coefficients of the series whose 24th power equals E_2(x)^5/E_10(x), where E_2(x) and E_10(x) are the Eisenstein series A006352 and A013974.

Original entry on oeis.org

1, 6, 7038, 2002644, 922569342, 380737463400, 175255606306116, 80315525064955440, 38028486993289854966, 18171889608389845598586, 8807723964899085718419480, 4305311468773791666900669828, 2122088430918938935321961736084
Offset: 0

Views

Author

Peter Bala, Feb 23 2021

Keywords

Comments

It is easy to see that E_2(x)^5/E_10(x) == 1 - 24*Sum_{k >= 1} (5*k - 11*k^9)*x^k/(1 - x^k) (mod 144), and also that the integer 5*k - 11*k^9 is always divisible by 6. Hence, E_2(x)^5/E_10(x) == 1 (mod 144). It follows from Heninger et al., p. 3, Corollary 2, that the series expansion of (E_2(x)^5/E_10(x))^(1/24) = 1 + 6*x + 7038*x^2 + 2002644*x^3 + 922569342*x^4 + ... has integer coefficients.

Crossrefs

Programs

  • Maple
    E(2,x)  := 1 -  24*add(k*x^k/(1-x^k),   k = 1..20):
    E(10,x) := 1 - 264*add(k^9*x^k/(1-x^k), k = 1..20):
    with(gfun): series((E(2,x)^5/E(10,x))^(1/24), x, 20):
    seriestolist(%);
Showing 1-9 of 9 results.