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 1-6 of 6 results.

A053282 Coefficients of the '10th-order' mock theta function psi(q).

Original entry on oeis.org

0, 1, 1, 2, 2, 2, 4, 4, 4, 6, 7, 8, 10, 11, 12, 16, 18, 20, 24, 26, 30, 36, 40, 44, 52, 58, 64, 74, 82, 91, 104, 116, 128, 144, 159, 176, 198, 218, 240, 268, 294, 324, 360, 394, 432, 478, 524, 572, 630, 688, 752, 826, 900, 980, 1072, 1168, 1270, 1386, 1505, 1634
Offset: 0

Views

Author

Dean Hickerson, Dec 19 1999

Keywords

Comments

Number of partitions (d1,d2,...,dm) of n such that 0 < d1/1 <= d2/2 <= ... <= dm/m. - Seiichi Manyama, Mar 17 2018

Examples

			From _Seiichi Manyama_, Mar 17 2018: (Start)
n | Partition (d1,d2,...,dm) | (d1/1, d2/2, ... , dm/m)
--+--------------------------+-------------------------
1 | (1)                      | (1)
2 | (2)                      | (2)
3 | (3)                      | (3)
  | (1, 2)                   | (1, 1)
4 | (4)                      | (4)
  | (1, 3)                   | (1, 3/2)
5 | (5)                      | (5)
  | (1, 4)                   | (1, 2)
6 | (6)                      | (6)
  | (1, 5)                   | (1, 5/2)
  | (2, 4)                   | (2, 2)
  | (1, 2, 3)                | (1, 1, 1)
7 | (7)                      | (7)
  | (1, 6)                   | (1, 3)
  | (2, 5)                   | (2, 5/2)
  | (1, 2, 4)                | (1, 1, 4/3)
8 | (8)                      | (8)
  | (1, 7)                   | (1, 7/2)
  | (2, 6)                   | (2, 3)
  | (1, 2, 5)                | (1, 1, 5/3)
9 | (9)                      | (9)
  | (1, 8)                   | (1, 4)
  | (2, 7)                   | (2, 7/2)
  | (3, 6)                   | (3, 3)
  | (1, 2, 6)                | (1, 1, 2)
  | (1, 3, 5)                | (1, 3/2, 5/3) (End)
		

References

  • Srinivasa Ramanujan, The Lost Notebook and Other Unpublished Papers, Narosa Publishing House, New Delhi, 1988, p. 9

Crossrefs

Other '10th-order' mock theta functions are at A053281, A053283, A053284.

Programs

  • Mathematica
    Series[Sum[q^((n+1)(n+2)/2)/Product[1-q^(2k+1), {k, 0, n}], {n, 0, 12}], {q, 0, 100}]
    nmax = 100; CoefficientList[Series[Sum[x^((k+1)*(k+2)/2) / Product[1-x^(2*j+1), {j, 0, k}], {k, 0, Floor[Sqrt[2*nmax]]}], {x, 0, nmax}], x] (* Vaclav Kotesovec, Jun 11 2019 *)

Formula

G.f.: psi(q) = Sum_{n >= 0} q^((n+1)(n+2)/2)/((1-q)(1-q^3)...(1-q^(2n+1))).
a(n) ~ exp(Pi*sqrt(n/5)) / (2*5^(1/4)*sqrt(phi*n)), where phi = A001622 = (1+sqrt(5))/2 is the golden ratio. - Vaclav Kotesovec, Jun 12 2019

A217094 Least index k such that A011540(k) >= 10^n.

Original entry on oeis.org

2, 2, 11, 182, 2621, 33572, 402131, 4619162, 51572441, 564151952, 6077367551, 64696307942, 682266771461, 7140400943132, 74263608488171, 768372476393522, 7915352287541681, 81238170587875112, 831143535290875991, 8480291817617883902, 86322626358560955101
Offset: 0

Views

Author

Hieronymus Fischer, Jan 23 2013

Keywords

Comments

For n>0 also index k such that A011540(k) = 10^n.
For n>1: A011540(a(n)) is the least number with n zero digits.
For n>0: a(n) - 1 is the number of numbers with <= n digits which contain the digit '0'. - Hieronymus Fischer, Dec 27 2013

Examples

			a(0) = 2, since A011540(2) = 10 >= 10^0.
a(1) = 2, since A011540(2) = 10 >= 10^1.
a(2) = 11, since A011540(11) = 100 >= 10^2, but A011540(10) = 90 < 10^2.
		

Crossrefs

Programs

  • Magma
    [2 +10^n -9^n -(9^n -1)/8: n in [0..50]]; // G. C. Greubel, Apr 18 2018
  • Mathematica
    LinearRecurrence[{20,-109,90},{2,2,11},30] (* Harvey P. Dale, Aug 02 2015 *)
  • PARI
    for(n=0,50, print1(2 +10^n -9^n -(9^n -1)/8, ", ")) \\ G. C. Greubel, Apr 18 2018
    

Formula

a(n+1) = 10*a(n) - 9*a(n-1) + 9*10^(n-1), n>0.
a(n) = 2 + 10^n - 9^n - (9^n - 1)/8.
A011540(a(n)) = 10^n, for n>0.
a(n) = A053283(A002452(n+1) - 1) - A002452(n+1) + 3, n>0.
a(n) = 10^n + 2 - A002452(n+1).
G.f.: (189*x^2 - 38*x + 2)/((1-x)*(1-9*x)*(1-10*x)).
a(n) = 1 + sum_{1<=k<=n} A229127(k), for n>0. - Hieronymus Fischer, Dec 27 2013

A053281 Coefficients of the '10th-order' mock theta function phi(q).

Original entry on oeis.org

1, 2, 2, 3, 4, 4, 6, 7, 8, 10, 12, 14, 16, 20, 22, 26, 31, 34, 40, 46, 52, 60, 68, 76, 87, 98, 110, 124, 140, 156, 174, 196, 216, 242, 270, 298, 332, 368, 406, 449, 496, 546, 602, 664, 728, 800, 880, 962, 1056, 1156, 1262, 1381, 1508, 1644, 1794, 1956, 2128
Offset: 0

Views

Author

Dean Hickerson, Dec 19 1999

Keywords

Comments

The alternating sum of the same series, namely phi(q) = Sum_{n>=0} (-1)^n q^(n(n+1)/2)/((1-q)(1-q^3)...(1-q^(2n+1))) = 1 + x^3 - x^7 - x^16 + x^24 + x^39 - x^51 - ..., where the exponents are given by 5n^2 +- 2n. See the Amer. Math. Monthly reference.

References

  • Srinivasa Ramanujan, The Lost Notebook and Other Unpublished Papers, Narosa Publishing House, New Delhi, 1988, p. 9.

Crossrefs

Other '10th-order' mock theta functions are at A053282, A053283, A053284.

Programs

  • Mathematica
    Series[Sum[q^(n(n+1)/2)/Product[1-q^(2k+1), {k, 0, n}], {n, 0, 13}], {q, 0, 100}]
    nmax = 100; CoefficientList[Series[Sum[x^(k*(k+1)/2) / Product[1-x^(2*j+1), {j, 0, k}], {k, 0, Floor[Sqrt[2*nmax]]}], {x, 0, nmax}], x] (* Vaclav Kotesovec, Jun 11 2019 *)

Formula

G.f.: phi(q) = Sum_{n >= 0} q^(n(n+1)/2)/((1-q)(1-q^3)...(1-q^(2n+1))).
a(n) ~ sqrt(phi) * exp(Pi*sqrt(n/5)) / (2*5^(1/4)*sqrt(n)), where phi = A001622 = (1+sqrt(5))/2 is the golden ratio. - Vaclav Kotesovec, Jun 12 2019

A053284 Coefficients of the '10th-order' mock theta function chi(q).

Original entry on oeis.org

0, 1, -1, 1, -2, 2, -1, 2, -3, 3, -3, 3, -4, 4, -4, 5, -6, 7, -6, 7, -9, 8, -8, 10, -12, 13, -13, 13, -16, 17, -16, 19, -21, 22, -23, 25, -28, 29, -30, 33, -37, 39, -39, 42, -48, 49, -50, 55, -60, 64, -66, 70, -77, 81, -82, 89, -97, 101, -105, 112, -121, 126, -131, 140, -151, 159, -163, 173, -187, 194, -202
Offset: 0

Views

Author

Dean Hickerson, Dec 19 1999

Keywords

References

  • Srinivasa Ramanujan, The Lost Notebook and Other Unpublished Papers, Narosa Publishing House, New Delhi, 1988, p. 9.

Crossrefs

Other '10th-order' mock theta functions are at A053281, A053282, A053283.

Programs

  • Mathematica
    Series[Sum[(-1)^n q^(n+1)^2/Product[1+q^k, {k, 1, 2n+1}], {n, 0, 9}], {q, 0, 100}]
    nmax = 100; CoefficientList[Series[Sum[(-1)^k * x^((k+1)^2)/Product[1+x^j, {j, 1, 2*k+1}], {k, 0, Floor[Sqrt[nmax]]}], {x, 0, nmax}], x] (* Vaclav Kotesovec, Jun 11 2019 *)

Formula

G.f.: chi(q) = Sum_{n >= 0} (-1)^n q^(n+1)^2/((1+q)(1+q^2)...(1+q^(2n+1))).
a(n) ~ -(-1)^n * sqrt(phi) * exp(Pi*sqrt(n/10)) / (2*5^(1/4)*sqrt(n)), where phi = A001622 = (1+sqrt(5))/2 is the golden ratio. - Vaclav Kotesovec, Jun 12 2019

A217110 Number of pandigital numbers with n places.

Original entry on oeis.org

0, 0, 0, 0, 0, 0, 0, 0, 0, 3265920, 179625600, 5568393600, 128432304000, 2458427811840, 41355201888000, 632788296940800, 9008498667168000, 121205358007493760, 1558813928579107200, 19326359087766057600, 232491479092720848000, 2727512837264447527680, 31331281164921975283200, 353549170783043484480000
Offset: 1

Views

Author

Hieronymus Fischer, Feb 13 2013

Keywords

Comments

The number of numbers between 10^(n-1) and 10^n which contain all decimal digits 0..9.
The ratio a(n)/(10^n-10^(n-1)) indicates the relative proportion of pandigital n-digit numbers compared to all n-digit numbers. Since that ratio converges to the limit 1 for n->oo this can be expressed for large numbers as follows (in a slightly popular manner): "Almost all numbers contain all decimal digits 0..9".
Example: a(n)/(10^n-10^(n-1)) = 0.99973439517775... for n = 100; in this case 99.9734...% of all 100-digit numbers contain all digits 0..9. Conversely, only the tiny proportion of 0.00026560482224... (< 0.03%) lacks one digit. That's astonishing! Intuitively, this is not what one would expect. In fact, for smaller numbers (with which most people are faced normally) the relative portion of numbers which missing at least one digit is significantly larger. Of course, for n < 10 the portion is 100%, and even for numbers with n = 10 or 20 digits the relative proportion of numbers which do not contain all digits 0..9 is 99.96371...% or 78.52626...%, respectively. The least number of digits for which the pandigital numbers hold the majority is 27. Here, the proportion of numbers which do not contain all digits is 48.03664...%. So one could bet that a randomly chosen number with >= 27 digits contains all digits.

Examples

			a(k) = 0 for k < 10 since there are no pandigital numbers with < 10 places, trivially.
a(10) = 9*9! since the first digit can be in the range 1..9 and for the following 9 digits there are 9, 8, 7, ..., 1 possible values.
		

Crossrefs

Formula

a(n) = 9*9!*S2(n,10), where the S2(n,10) are the Stirling numbers of the second kind (cf. triangle A008277).
Asymptotic behavior: Limit_{n->oo} a(n)/10^n = 9/10.
G.f.: g(x) = 9*9!*x^10/(Product_{j=1..10} (1-jx)).
E.g.f. g(x) = (9/10) * (e^x - 1)^10.

A217111 Number of pandigital numbers <= 10^n.

Original entry on oeis.org

0, 0, 0, 0, 0, 0, 0, 0, 0, 3265920, 182891520, 5751285120, 134183589120, 2592611400960, 43947813288960, 676736110229760, 9685234777397760, 130890592784891520, 1689704521363998720, 21016063609130056320, 253507542701850904320, 2981020379966298432000
Offset: 1

Views

Author

Hieronymus Fischer, Feb 13 2013

Keywords

Comments

The number of numbers with <= n digits which contain all decimal digits 0..9.
The ratio a(n)/10^n indicates the relative proportion of pandigital numbers <= 10^n compared to all numbers <= 10^n. Since that ratio converges to the limit 1 for n -> oo this can be expressed for large numbers as follows (in a slightly popular manner): "Almost all numbers contain all decimal digits 0..9".
Example: a(n)/10^n = 0. 99973107526479... for n = 100; in this case 99.9731...% of all numbers <= 10^100 contain all digits 0..9. Conversely, only the tiny proportion of 0.000268924735210... (< 0.03%) lacks at least one digit. That's astonishing! Intuitively, this is not what one would expect. In fact, for smaller numbers (with which most people are faced normally) the relative portion of numbers which missing at least one digit is significantly larger. Of course, for n < 10 the portion is 100%, and even for numbers <= 10^10 or <= 10^20 the relative proportion of numbers which do not contain all digits 0..9 is 99.96734...% or 78.98393...%, respectively. 10^27 is the least power of 10 such that the pandigital numbers hold the majority. Here, the proportion of pandigital numbers among all numbers <= 10^27 is 51.50961...%. So one could bet that a randomly chosen number <= 10^27 contains all digits.
Partial sums of A217110.

Examples

			a(k) = 0, for k < 10 since there are no pandigital numbers <= 10^9, trivially.
a(10) = 9*9!, since the first digit can be in the range 1..9 and for the following 9 digits there are 9, 8, 7, ..., 1 possible values.
		

Crossrefs

Programs

  • Mathematica
    3265920 Accumulate[StirlingS2[Range[25],10]] (* Harvey P. Dale, Oct 16 2022 *)

Formula

a(n) = 9*9!*Sum_{j=1..n} S2(j,10), where the S2(j,10) are the Stirling numbers of the second kind (cf. triangle A008277).
Asymptotic behavior:
Limit_{n->oo} a(n)/10^n = 1.
G.f.: g(x) = 9*9!*x^10/((1-x)*Product_{j=1..10} (1-jx)).
Showing 1-6 of 6 results.