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 1-10 of 15 results. Next

A364173 a(n) = (9*n)!*(2*n)!*(3*n/2)!/((9*n/2)!*(4*n)!*(3*n)!*n!).

Original entry on oeis.org

1, 128, 43758, 17039360, 7012604550, 2976412336128, 1288415796384780, 565399665327996928, 250622090889055155270, 111950839825145979207680, 50312973039218473430585508, 22723567527558510746926055424, 10304958075870392958137083227804
Offset: 0

Views

Author

Peter Bala, Jul 13 2023

Keywords

Comments

A295440, defined by A295440(n) = (18*n)!*(4*n)!*(3*n)! / ((9*n)!*(8*n)!*(6*n)!*(2*n)!), is one of the 52 sporadic integral factorial ratio sequences of height 1 found by V. I. Vasyunin (see Bober, Table 2, Entry 10). Here we are essentially considering the sequence {A295440(n/2) : n >= 0}. Fractional factorials are defined in terms of the gamma function; for example, (3*n/2)! := Gamma(1 + 3*n/2).
This sequence is only conjecturally an integer sequence.
Conjecture: the supercongruences a(n*p^r) == a(n*p^(r-1)) (mod p^(3*r)) hold for all primes p >= 5 and all positive integers n and r.

Crossrefs

Programs

  • Maple
    seq( simplify((9*n)!*(2*n)!*(3*n/2)!/((9*n/2)!*(4*n)!*(3*n)!*n!)) , n = 0..15);

Formula

a(n) ~ c^n * 1/sqrt(4*Pi*n), where c = (3^7)/(2^3) * sqrt(3) = 473.4993895191418....
a(n) = 108*(9*n - 1)*(9*n - 5)*(9*n - 7)*(9*n - 11)*(9*n - 13)*(9*n - 17)/(n*(n - 1)*(4*n - 1)*(4*n - 3)*(4*n - 5)*(4*n - 7))*a(n-2) for n >= 2 with a(0) = 1 and a(1) = 128.

A363249 Leading digit of 9^n.

Original entry on oeis.org

1, 9, 8, 7, 6, 5, 5, 4, 4, 3, 3, 3, 2, 2, 2, 2, 1, 1, 1, 1, 1, 1, 9, 8, 7, 7, 6, 5, 5, 4, 4, 3, 3, 3, 2, 2, 2, 2, 1, 1, 1, 1, 1, 1, 9, 8, 7, 7, 6, 5, 5, 4, 4, 3, 3, 3, 2, 2, 2, 1, 1, 1, 1, 1, 1, 1, 9, 8, 7, 6, 6, 5, 5, 4, 4, 3, 3, 2, 2, 2, 2, 1, 1, 1, 1, 1, 1, 1, 9, 8, 7, 6, 6, 5, 4, 4, 4, 3, 3, 2, 2
Offset: 0

Views

Author

Seiichi Manyama, Jul 15 2023

Keywords

References

  • He, Xinwei; Hildebrand, A J; Li, Yuchen; Zhang, Yunyi, Complexity of Leading Digit Sequences, Discrete Mathematics and Theoretical Computer Science; 22 (2020), #14.

Crossrefs

Programs

  • Mathematica
    a[n_] := IntegerDigits[9^n][[1]]; Array[a, 100, 0] (* Amiram Eldar, Jul 15 2023 *)
  • PARI
    a(n) = digits(9^n)[1];

Formula

a(n) = A000030(A001019(n)).
a(n) = A060956(2*n).

A362871 Leading digit of 6^n.

Original entry on oeis.org

1, 6, 3, 2, 1, 7, 4, 2, 1, 1, 6, 3, 2, 1, 7, 4, 2, 1, 1, 6, 3, 2, 1, 7, 4, 2, 1, 1, 6, 3, 2, 1, 7, 4, 2, 1, 1, 6, 3, 2, 1, 8, 4, 2, 1, 1, 6, 3, 2, 1, 8, 4, 2, 1, 1, 6, 3, 2, 1, 8, 4, 2, 1, 1, 6, 3, 2, 1, 8, 4, 2, 1, 1, 6, 3, 2, 1, 8, 4, 2, 1, 1, 6, 3, 2, 1, 8, 5, 3, 1, 1, 6, 3, 2, 1, 8, 5, 3, 1, 1, 6
Offset: 0

Views

Author

Seiichi Manyama, Jul 15 2023

Keywords

Crossrefs

Programs

  • Mathematica
    a[n_] := IntegerDigits[6^n][[1]]; Array[a, 100, 0] (* Amiram Eldar, Jul 15 2023 *)
  • PARI
    a(n) = digits(6^n)[1];

Formula

a(n) = A000030(A000400(n)).

A363093 Leading digit of 7^n.

Original entry on oeis.org

1, 7, 4, 3, 2, 1, 1, 8, 5, 4, 2, 1, 1, 9, 6, 4, 3, 2, 1, 1, 7, 5, 3, 2, 1, 1, 9, 6, 4, 3, 2, 1, 1, 7, 5, 3, 2, 1, 1, 9, 6, 4, 3, 2, 1, 1, 7, 5, 3, 2, 1, 1, 8, 6, 4, 3, 2, 1, 1, 7, 5, 3, 2, 1, 1, 8, 5, 4, 2, 2, 1, 1, 7, 4, 3, 2, 1, 1, 8, 5, 4, 2, 1, 1, 9, 6, 4, 3, 2, 1, 1, 8, 5, 3, 2, 1, 1, 9, 6, 4, 3
Offset: 0

Views

Author

Seiichi Manyama, Jul 15 2023

Keywords

Crossrefs

Programs

  • Mathematica
    a[n_] := IntegerDigits[7^n][[1]]; Array[a, 100, 0] (* Amiram Eldar, Jul 15 2023 *)
  • PARI
    a(n) = digits(7^n)[1];

Formula

a(n) = A000030(A000420(n)).

A364172 a(n) = (6*n)!*(n/3)!/((3*n)!*(2*n)!*(4*n/3)!).

Original entry on oeis.org

1, 45, 6237, 1021020, 178719453, 32427545670, 6016814703900, 1133540594837892, 215925912619400925, 41477110789150966020, 8019784929635201045862, 1558875476359831844951100, 304331361887290342345862940, 59629409730107012112361325820
Offset: 0

Views

Author

Peter Bala, Jul 12 2023

Keywords

Comments

Fractional factorials are defined in terms of the gamma function; for example, (n/3)! := Gamma(n/3 + 1).
Given two sequences of numbers c = (c_1, c_2, ..., c_K) and d = (d_1, d_2, ..., d_L) where c_1 + ... + c_K = d_1 + ... + d_L we can define the factorial ratio sequence u_n(c, d) = (c_1*n)!*(c_2*n)!* ... *(c_K*n)!/ ( (d_1*n)!*(d_2*n)!* ... *(d_L*n)! ) and ask whether it is integral for all n >= 0. The integer L - K is called the height of the sequence. Bober completed the classification of integral factorial ratio sequences of height 1. For a list of the 52 sporadic integral factorial ratio sequences see A295431.
It is usually assumed that the c's and d's are integers but here we allow for some of the c's and d's to be rational numbers.
A295437, defined by A295437(n) = (18*n)!*n! / ((9*n)!*(6*n)!*(4*n)!) is one of the 52 sporadic integral factorial ratio sequences of height 1 found by V. I. Vasyunin (see Bober, Table 2, Entry 7). Here we are essentially considering the sequence {A295437(n/3) : n >= 0}. This sequence is only conjecturally an integer sequence.
Conjecture: the supercongruences a(n*p^r) == a(n*p^(r-1)) (mod p^(3*r)) hold for all primes p >= 5 and all positive integers n and r.

Crossrefs

Programs

  • Maple
    seq( simplify((6*n)!*(n/3)!/((3*n)!*(2*n)!*(4*n/3)!)), n = 0..15);
  • Mathematica
    Table[Product[36*(6*k - 5)*(6*k - 1)/(k*(3*k + n)), {k, 1, n}], {n, 0, 20}] (*  Vaclav Kotesovec, Jul 13 2023 *)

Formula

a(n) ~ 2^(4*n/3 - 3/2) * 3^(4*n) / sqrt(Pi*n). - Vaclav Kotesovec, Jul 13 2023
a(n) = 5832*(6*n - 1)*(6*n - 5)*(6*n - 7)*(6*n - 11)*(6*n - 13)*(6*n - 17)/(n*(n - 1)*(n - 2)*(2*n - 3)*(4*n - 3)*(4*n - 9))*a(n-3) for n >= 3 with a(0) = 1, a(1) = 45 and a(2) = 6237.

A364183 a(n) = (12*n)!*(2*n)!*(n/2)!/((6*n)!*(4*n)!*(7*n/2)!*n!).

Original entry on oeis.org

1, 4224, 76488984, 1626105446400, 36856530424884600, 864687003650148532224, 20728451893251973782071160, 504292670666772382512278667264, 12401082728528113445556802226795640, 307453669544695584297743425538327838720, 7671567513095586883562392061857092727662984
Offset: 0

Views

Author

Peter Bala, Jul 13 2023

Keywords

Comments

A295479, defined by A295479(n) = (24*n)!*(4*n)!*n! / ((12*n)!*(8*n)!*(7*n)!*(2*n)!), is one of the 52 sporadic integral factorial ratio sequences of height 1 found by V. I. Vasyunin (see Bober, Table 2, Entry 49). Here we are essentially considering the sequence {A295479(n/2) : n >= 0}. Fractional factorials are defined in terms of the gamma function; for example, (7*n/2)! := Gamma(1 + 7*n/2).
This sequence is only conjecturally an integer sequence.
Conjecture: the supercongruences a(n*p^r) == a(n*p^(r-1)) (mod p^(3*r)) hold for all primes p >= 5 and all positive integers n and r.

Crossrefs

Programs

  • Maple
    seq( simplify((12*n)!*(2*n)!*(n/2)!/((6*n)!*(4*n)!*(7*n/2)!*n!)), n = 0..15);

Formula

a(n) ~ c^n * 1/sqrt(14*Pi*n), where c = (2^15)*(3^6)/(7^4) * sqrt(7).
a(n) = 1327104*(12*n - 1)*(12*n - 5)*(12*n - 7)*(12*n - 11)*(12*n - 13)*(12*n - 17)*(12*n - 19)*(12*n - 23)/(7*n*(n - 1)*(7*n - 2)*(7*n - 4)*(7*n - 6)*(7*n - 8)*(7*n - 10)*(7*n - 12))*a(n-2) with a(0) = 1 and a(1) = 4224.

A364174 a(n) = (9*n)!*(5*n/2)!*(3*n/2)!/((5*n)!*(9*n/2)!*(3*n)!*(n/2)!).

Original entry on oeis.org

1, 48, 4862, 549120, 65132550, 7945986048, 987291797996, 124259864002560, 15789207515217990, 2021092963752345600, 260227401685879140612, 33665720694993527504896, 4372592850984736084611996, 569819472537519480058675200, 74468439316740019538310543000
Offset: 0

Views

Author

Peter Bala, Jul 13 2023

Keywords

Comments

A295442, defined by A295442(n) = (18*n)!*(5*n)!*(3*n)!/((10*n)!*(9*n)!*(6*n)!*n!), is one of the 52 sporadic integral factorial ratio sequences of height 1 found by V. I. Vasyunin (see Bober, Table 2, Entry 12). Here we are essentially considering the sequence {A295442(n/2) : n >= 0}. Fractional factorials are defined in terms of the gamma function; for example, (3*n/2)! := Gamma(1 + 3*n/2).
This sequence is only conjecturally an integer sequence.
Conjecture: the supercongruences a(n*p^r) == a(n*p^(r-1)) (mod p^(3*r)) hold for all primes p >= 5 and all positive integers n and r.

Crossrefs

Programs

  • Maple
    seq( simplify((9*n)!*(5*n/2)!*(3*n/2)!/((5*n)!*(9*n/2)!*(3*n)!*(n/2)!)), n = 0..15)

Formula

a(n) ~ c^n * 1/sqrt(2*Pi*n), where c = 2*(3^7)/(5^3) * sqrt(15) = 135.5234332504899....
a(n) = 108*(9*n - 1)*(9*n - 5)*(9*n - 7)*(9*n - 11)*(9*n - 13)*(9*n - 17)/(5*n*(n - 1)*(5*n - 1)*(5*n - 3)*(5*n - 7)*(5*n - 9))*a(n-2) for n >= 2 with a(0) = 1 and a(1) = 48.

A364175 a(n) = (6*n)!*(2*n/3)!/((3*n)!*(2*n)!*(5*n/3)!).

Original entry on oeis.org

1, 36, 3564, 408408, 49697388, 6249195036, 802241960520, 104466877291260, 13746018177013356, 1823169705017624880, 243331037661693468564, 32641262295291161362656, 4396944340992842923469640, 594371374049863341847620936, 80586283761263090599592845140
Offset: 0

Views

Author

Peter Bala, Jul 13 2023

Keywords

Comments

A295445, defined by A295445(n) = (18*n)!*(2*n)! / ((9*n)!*(6*n)!*(5*n)!), is one of the 52 sporadic integral factorial ratio sequences of height 1 found by V. I. Vasyunin (see Bober, Table 2, Entry 15). Here we are essentially considering the sequence {A295445(n/3) : n >= 0}. Fractional factorials are defined in terms of the gamma function; for example, (2*n/3)! := Gamma(1 + 2*n/3).
This sequence is only conjecturally an integer sequence.
Conjecture: the supercongruences a(n*p^r) == a(n*p^(r-1)) (mod p^(3*r)) hold for all primes p >= 5 and all positive integers n and r.

Crossrefs

Programs

  • Maple
    seq( simplify((6*n)!*(2*n/3)!/((3*n)!*(2*n)!*(5*n/3)!)), n = 0..15);

Formula

a(n) ~ c^n * 1/sqrt(5*Pi*n) where c = (1296/25)*20^(1/3) = 140.7154092442799....
a(n) = 93312*(2*n - 3)*(6*n - 1)*(6*n - 5)*(6*n - 7)*(6*n - 11)*(6*n - 13)*(6*n - 17)/(5*n*(n - 1)*(n - 2)*(5*n - 3)*(5*n - 6)*(5*n - 9)*(5*n - 12))*a(n-3) with a(0) = 1, a(1) = 36 and a(2) = 3564.

A364176 a(n) = (15*n)!*(5*n/2)!*(2*n)!/((15*n/2)!*(6*n)!*(5*n)!*n!).

Original entry on oeis.org

1, 7168, 168043980, 4488240824320, 126694219977836700, 3688258943632086663168, 109504706026534324525391988, 3295939064766794222800490987520, 100204869963549181630558779565943580, 3070025447039504554088467623457608171520, 94632263448378916462441320194245442445186480
Offset: 0

Views

Author

Peter Bala, Jul 13 2023

Keywords

Comments

A295456, defined by A295456(n) = (30*n)!*(5*n)!*(4*n)! / ((15*n)!*(12*n)!*(10*n)!*(2*n)!), is one of the 52 sporadic integral factorial ratio sequences of height 1 found by V. I. Vasyunin (see Bober, Table 2, Entry 26). Here we are essentially considering the sequence {A295456(n/2) : n >= 0}. Fractional factorials are defined in terms of the gamma function; for example, (5*n/2)! := Gamma(1 + 5*n/2).
This sequence is only conjecturally an integer sequence.
Conjecture: the supercongruences a(n*p^r) == a(n*p^(r-1)) (mod p^(3*r)) hold for all primes p >= 5 and all positive integers n and r.

Crossrefs

Programs

  • Maple
    seq( simplify((15*n)!*(5*n/2)!*(2*n)!/((15*n/2)!*(6*n)!*(5*n)!*n!)), n = 0..15)

Formula

a(n) ~ c^n * 1/sqrt(6*Pi*n), where c = 18750*sqrt(3).
a(n) = 4800*(15*n - 1)*(15*n - 7)*(15*n - 11)*(15*n - 13)*(15*n - 17)*(15*n - 19)*(15*n - 23)*(15*n - 29)/(n*(n - 1)*(3*n - 2)*(3*n - 4)*(6*n - 1)*(6*n - 5)*(6*n - 7)*(6*n - 11))*a(n-2) with a(0) = 1 and a(1) = 7168.

A364177 a(n) = (15*n)!*(5*n/2)!*(2*n)!/((15*n/2)!*(5*n)!*(4*n)!*(3*n)!).

Original entry on oeis.org

1, 35840, 5545451340, 991901222174720, 188242272043069768860, 36901030731039027064995840, 7383354803839076831124554790900, 1498315221854950975184507333477662720, 307213802011837003346320048243705086348060
Offset: 0

Views

Author

Peter Bala, Jul 13 2023

Keywords

Comments

A295458, defined by A295458(n) = (30*n)!*(5*n)!*(4*n)! / ((15*n)!*(10*n)!*(8*n)!*(6*n)!), is one of the 52 sporadic integral factorial ratio sequences of height 1 found by V. I. Vasyunin (see Bober, Table 2, Entry 28). Here we are essentially considering the sequence {A295458(n/2) : n >= 0}. Fractional factorials are defined in terms of the gamma function; for example, (5*n/2)! := Gamma(1 + 5*n/2).
This sequence is only conjecturally an integer sequence.
Conjecture: the supercongruences a(n*p^r) == a(n*p^(r-1)) (mod p^(3*r)) hold for all primes p >= 5 and all positive integers n and r.

Crossrefs

Programs

  • Maple
    seq( simplify((15*n)!*(5*n/2)!*(2*n)!/((15*n/2)!*(5*n)!*(4*n)!*(3*n)!)), n = 0..15);

Formula

a(n) ~ c^n * 1/sqrt(12*Pi*n), where c = (3^4)*(5^5) * sqrt(3)/2.
a(n) = 43200*(15*n - 1)*(15*n - 7)*(15*n - 11)*(15*n - 13)*(15*n - 17)*(15*n - 19)*(15*n - 23)*(15*n - 29)/(n*(n - 1)*(3*n - 2)*(3*n - 4)*(4*n - 1)*(4*n - 3)*(4*n - 5)*(4*n - 7))*a(n-2) with a(0) = 1 and a(1) = 35840.
Showing 1-10 of 15 results. Next