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.

A091137 The Hirzebruch numbers. a(n) = Product_{2 <= p <= n+1, p prime} p^floor(n / (p - 1)).

Original entry on oeis.org

1, 2, 12, 24, 720, 1440, 60480, 120960, 3628800, 7257600, 479001600, 958003200, 2615348736000, 5230697472000, 31384184832000, 62768369664000, 32011868528640000, 64023737057280000, 51090942171709440000, 102181884343418880000, 33720021833328230400000, 67440043666656460800000
Offset: 0

Views

Author

Henry Bottomley, Dec 19 2003

Keywords

Comments

Largest number m such that number of times m divides k! is almost k/n for large k, i.e., largest m with A090624(m) = n.
This is always a relatively small multiple of n!, since the multiplicity with which a prime p divides n! is always <= n/(p-1); it is equal to floor(n/(p-1)) at least when n is a power of p. - Franklin T. Adams-Watters, May 31 2010
At least for most small n, a(n) = A002790(n) * n!; the first difference is n=15. It appears that A002790(n) * n! always divides a(n).
Conjecture: The denominators of the series reversion of the sequence with e.g.f. Polylog(2,x). - Benedict W. J. Irwin, Jan 05 2017
Not only is a(n) divisible by n!; a(n) is divisible by (n + 1)! as has been observed by Bedhouche and Bakir (see links and A363596). - Hal M. Switkay, Aug 15 2025

Examples

			Let n = 4. The partitions of 4 are [[4], [3, 1], [2, 2], [2, 1, 1], [1, 1, 1, 1]]. Thus a(4) = lcm([5, 4*2, 3*3, 3*2*2, 2*2*2*2]) = 720.
		

References

  • P. Curtz, Integration numérique ..., Note 12, C.C.S.A., Arcueil, 1969; see pp. 36, 56.
  • F. Hirzebruch, Topological Methods in Algebraic Geometry, Springer, 3rd. ed., 1966; Lemma 1.7.3, p. 14. [From N. J. A. Sloane, Sep 06 2010]

Crossrefs

Starts similarly to A002207 especially for even n and all values of A002207 seen so far seem to divide a(n).

Programs

  • Maple
    A091137 := proc(n) local a,i,p ; a := 1 ; for i from 1 do p := ithprime(i) ; if p > n+1 then break; fi; a := a*p^floor(n/(p-1)) ; od: a ; end:
    seq(A091137(n), n = 0..47); # R. J. Mathar, Feb 23 2009
  • Mathematica
    A027760[n_] := Product[d, {d, Select[ Divisors[n] + 1, PrimeQ]}]; a[n_] := a[n] = A027760[n]*a[n-1]; a[0] = 1; Table[ a[n], {n, 0, 18}] (* Jean-François Alcover, Oct 04 2011 *)
  • PARI
    a(n) = local(r); r=1; forprime(p=2, n+1, r*=p^(n\(p-1))); r
    \\ Franklin T. Adams-Watters, May 31 2010
    
  • Python
    from math import prod
    from sympy import primerange
    def A091137(n): return prod(p**(n//(p-1)) for p in primerange(n+2))
    # Chai Wah Wu, Apr 28 2023
    
  • SageMath
    def a(n): return lcm(product(r + 1 for r in p) for p in Partitions(n))
    # Or, more efficient:
    from functools import cache
    @cache
    def a_rec(n):
        if n == 0: return 1
        p = mul(s for s in map(lambda i: i + 1, divisors(n)) if is_prime(s))
        return p * a_rec(n - 1)
    print([a_rec(n) for n in range(22)]) # Peter Luschny, Dec 12 2023

Formula

a(n) = Product_p {p prime} p^floor(n/(p-1)).
a(2n+1) = 2*a(2n).
a(n+1) = A027760(n+1)*a(n). - Paul Curtz, Aug 01 2008
From Peter Luschny, Dec 11 2023: (Start)
a(n) = lcm_{p in P(n)} Product_{r in p}(r + 1), where P(n) are the partitions of n.
a(n) = lcm(A238963row(n)).
a(n) = A368116(1, n), seen as the lcm of the product of the 1-shifted partitions.
a(n) = A368093(1, n), seen as the cumulative product of the Clausen numbers A160014(1, n). (End)
a(n) = lcm({k: A275314(k) = n+1}). - Hal M. Switkay, Aug 13 2025
a(n) = (n + 1)! * A363596(n). - Hal M. Switkay, Aug 15 2025

Extensions

New name using a formula of the author by Peter Luschny, Dec 11 2023

A099375 Sequence matrix for odd numbers.

Original entry on oeis.org

1, 3, 1, 5, 3, 1, 7, 5, 3, 1, 9, 7, 5, 3, 1, 11, 9, 7, 5, 3, 1, 13, 11, 9, 7, 5, 3, 1, 15, 13, 11, 9, 7, 5, 3, 1, 17, 15, 13, 11, 9, 7, 5, 3, 1, 19, 17, 15, 13, 11, 9, 7, 5, 3, 1, 21, 19, 17, 15, 13, 11, 9, 7, 5, 3, 1, 23, 21, 19, 17, 15, 13, 11, 9, 7, 5, 3, 1, 25, 23, 21, 19, 17, 15, 13, 11, 9
Offset: 0

Views

Author

Paul Barry, Jan 22 2005

Keywords

Comments

Riordan array ((1+x)/(1-x)^2, x).
Inverse matrix is A101038.
Row sums yield (n+1)^2.
Diagonal sums yield sum{k=0..floor(n/2),2(n-2k)+1}=C(n+2,2)=A000217(n+1). Note that sum{k=0..n,2(n-2k)+1}=n+1.
From Paul Curtz, Sep 25 2011. (Start)
Consider from A187870(n-2) and A171080(n)
1 + 1/3 - 4/45 + 44/945 - 428/14175 =1/(1 -1/3 +1/5 -1/7 ..= Pi/4)=4/Pi.
For c(0)=-1, c(1)=1/3, c(2)=4/45, c(3)=44/945, c(4)=428/14175,
c(0)/3 + c(1)=0,
c(0)/5 + c(1)/3 + c(2)=0,
c(0)/7 + c(1)/5 + c(2)/3 + c(3)=0.
Hence a(n+1). Numbers are
-1/3 + 1/3, 1=1,
-1/5 + 1/9 + 4/45, 4=9-5,
-1/7 + 1/15 + 4/135 + 44/945 44=135-63-28. (End)
T(n,k) = A158405(n+1,n+1-k), 1<=k<=n. [Reinhard Zumkeller, Mar 31 2012]
From Peter Bala, Jul 22 2014: (Start)
Call this array M and for k = 0,1,2,... define M(k) to be the lower unit triangular block array
/I_k 0\
\ 0 M/
having the k x k identity matrix I_k as the upper left block; in particular, M(0) = M. The infinite matrix product M(0)*M(1)*M(2)*..., which is clearly well-defined, is equal to A208904. (End)

Examples

			Rows start
1;
3,1;
5,3,1;
7,5,3,1;
9,7,5,3,1;
11,9,7,5,3,1;
13,11,9,7,5,3,1;
		

Crossrefs

Programs

  • Haskell
    a099375 n k = a099375_row n !! k
    a099375_row n = a099375_tabl !! n
    a099375_tabl = iterate (\xs -> (head xs + 2) : xs) [1]
    -- Reinhard Zumkeller, Mar 31 2012

Formula

Number triangle T(n, k)=if(k<=n, 2(n-k)+1, 0)=binomial(2(n-k)+1, 2(n-k))
a(n)=2*A004736(n)-1; a(n)=2*((t*t+3*t+4)/2-n)-1, where t=floor((-1+sqrt(8*n-7))/2). - Boris Putievskiy, Feb 08 2013

A165636 a(n) = A091137(n)/2^n.

Original entry on oeis.org

1, 1, 3, 3, 45, 45, 945, 945, 14175, 14175, 467775, 467775, 638512875, 638512875, 1915538625, 1915538625, 488462349375, 488462349375, 194896477400625, 194896477400625, 32157918771103125, 32157918771103125, 2218896395206115625, 2218896395206115625, 3028793579456347828125, 3028793579456347828125, 9086380738369043484375
Offset: 0

Views

Author

Paul Curtz, Sep 23 2009

Keywords

Crossrefs

Programs

  • Maple
    A091137 := proc(n) local a, i, p ; a := 1 ; for i from 1 do p := ithprime(i) ; if p > n+1 then break; end if; a := a*p^floor(n/(p-1)) ; end do: a ; end proc:
    A165636 := proc(n) A091137(n)/2^n ; end proc: # R. J. Mathar, Jul 07 2011
  • PARI
    a(n)=my(p=primes(primepi(n+1)));prod(i=1,#p,p[i]^(n\(p[i]-1)))>>n \\ Charles R Greathouse IV, Jul 07 2011

Formula

a(n+1) = a(n)* A141459(n+1).

A368093 Cumulative products of the generalized Clausen numbers. Array read by ascending antidiagonals.

Original entry on oeis.org

1, 1, 1, 1, 2, 2, 1, 3, 12, 6, 1, 1, 9, 24, 12, 1, 5, 5, 135, 720, 60, 1, 1, 25, 5, 405, 1440, 360, 1, 7, 7, 875, 175, 8505, 60480, 2520, 1, 1, 49, 7, 4375, 175, 127575, 120960, 5040, 1, 1, 1, 343, 49, 21875, 875, 382725, 3628800, 15120
Offset: 0

Views

Author

Peter Luschny, Dec 12 2023

Keywords

Comments

A160014 are the generalized Clausen numbers, for m = 0 the formula computes the cumulative radical A048803, and for m = 1 the Hirzebruch numbers A091137.

Examples

			Array A(m, n) starts:
  [0] 1, 1,  2,   6,   12,     60,     360,      2520, ...  A048803
  [1] 1, 2, 12,  24,  720,   1440,   60480,    120960, ...  A091137
  [2] 1, 3,  9, 135,  405,   8505,  127575,    382725, ...  A368092
  [3] 1, 1,  5,   5,  175,    175,     875,       875, ...
  [4] 1, 5, 25, 875, 4375,  21875,  765625,  42109375, ...
  [5] 1, 1,  7,   7,   49,     49,    3773,      3773, ...
  [6] 1, 7, 49, 343, 2401, 184877, 1294139, 117766649, ...
  [7] 1, 1,  1,   1,   11,     11,     143,       143, ...
  [8] 1, 1,  1,  11,   11,    143,    1573,      1573, ...
  [9] 1, 1, 11,  11, 1573,   1573,   17303,     17303, ...
		

Crossrefs

Cf. A160014, A048803 (m=0), A091137 (m=1), A368092 (m=2).

Programs

  • SageMath
    from functools import cache
    def Clausen(n, k):
        return mul(s for s in map(lambda i: i+n, divisors(k)) if is_prime(s))
    @cache
    def CumProdClausen(m, n):
        return Clausen(m, n) * CumProdClausen(m, n - 1) if n > 0 else 1
    for m in range(10): print([m], [CumProdClausen(m, n) for n in range(8)])

Formula

A(m, n) = A160014(m, n) * A(m, n - 1) for n > 0 and A(m, 0) = 1.

A237111 Triangle read by rows: numerators of coefficients of the Hirzebruch L-polynomials L_n expressing the signature of a 4n-dimensional manifold in terms of its Pontrjagin numbers (as in Hirzebruch Signature Theorem).

Original entry on oeis.org

1, 7, -1, 62, -13, 2, 381, -71, -19, 22, -3, 5110, -919, -336, 237, 127, -83, 10, 2828954, -503904, -159287, 122523, -40247, 86901, -33863, 8718, -27635, 12842, -1382, 3440220, -611266, -185150, 146256, -62274, 88137, -37290, 22027, 16696, -39341, 10692, -7978, 11880, -4322, 420
Offset: 1

Views

Author

Carl McTague, Feb 03 2014

Keywords

Comments

The monomials of each polynomial L_n have been written in descending lexicographic order (where the exponent of p_1 is considered less significant than that of p_2, etc.) and over a common denominator. These denominators follow A171080.

Examples

			L_1 = p_1/3.
L_2 = (7 p_2-p_1^2)/45.
L_3 = (62 p_3-13 p_2 p_1+2 p_1^3)/945.
L_4 = (381 p_4-71 p_3 p_1-19 p_2^2+22 p_2 p_1^2-3 p_1^4)/14175.
L_5 = (5110 p_5-919 p_4 p_1-336 p_3 p_2+237 p_3 p_1^2+127 p_2^2 p_1-83 p_2 p_1^3+10 p_1^5)/467775.
The denominators 3,45,945,14175,467775 follow A171080.
The sequence of numerators thus begins 1,7,-1,62,-13,2,381,-71,-19,22,-3,5110,-919,-336,237,127,-83,10,...
		

References

  • F. Hirzebruch, Topological methods in algebraic geometry. Third enlarged edition. Die Grundlehren der Mathematischen Wissenschaften, Band 131 Springer-Verlag New York, Inc., New York 1966, p. 12.
  • F. Hirzebruch, The signature theorem: reminiscences and recreation. Prospects in mathematics (Proc. Sympos., Princeton Univ., Princeton, N.J., 1970), pp. 3-31. Ann. of Math. Studies, No. 70, Princeton Univ. Press, Princeton, N.J. 1971.

Crossrefs

Cf. A171080.

Programs

  • Mathematica
    K[Q_,n_Integer] := Module[{z,x},
       SymmetricReduction[
          SeriesCoefficient[
           Product[ComposeSeries[Series[Q[z],{z,0,n}],
             Series[Subscript[x,i]z,{z,0,n}]],{i,1,n}],n],
          Table[Subscript[x,i],{i,1,n}],
          Table[Subscript[p,i],{i,1,n}]][[1]] // FactorTerms];
    Table[K[Sqrt[#]/Tanh[Sqrt[#]]&, n],{n,1,5}]
Showing 1-5 of 5 results.