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 1-4 of 4 results.

A352346 Common terms between A061078 and A061077.

Original entry on oeis.org

26, 52, 148, 280, 320, 454, 1150, 1480, 8000, 41650, 80300, 165656, 166088, 614900, 2353700, 2859460, 28233200, 66130400, 68941640, 85717240, 107300320, 131507080, 155478800, 207666520, 1426680920, 1824596800, 2468014900, 2475648820, 5342351060, 5355218900, 5857281500, 8550475900, 36025361120
Offset: 1

Views

Author

Luca Onnis, Mar 12 2022

Keywords

Comments

Smarandache's conjecture: there are infinitely many terms.
This is a subsequence of A061076.

Examples

			26 is a term of this sequence, in fact:
26 = 1+3+5+7+9+1*1 (A061077(6)=26);
26 = 2+4+6+8+1*0+1*2+1*4 (A061078(7)=26).

References

  • A. Murthy, Smarandache friendly numbers and a few more sequences, Smarandache Notions Journal, Vol. 12, No. 1-2-3, Spring 2001. Page 267

Links

Crossrefs

Cf. A061076, A061077, A061078.

Programs

  • Mathematica
    Intersection[Accumulate[Times @@@ IntegerDigits[Range[2, 10000000, 2]]],
Accumulate[Times @@@ IntegerDigits[Range[1, 10000000, 2]]]]
  • Python
    from math import prod
from itertools import islice
def A352346_gen(): # generator of terms
    n1, m1, n2, m2 = 1, 1, 2, 2
    while True:
        if m1 == m2:
            yield m1
        k = 0
        while k == 0:
            n1 += 2
            m1 += (k := prod(int(d) for d in str(n1)))
        while m2 < m1:
            n2 += 2
            m2 += prod(int(d) for d in str(n2))
A352346_list = list(islice(A352346_gen(),20)) # Chai Wah Wu, Mar 21 2022

A350744 Numbers m such that A061078(m)/A061077(m) = 4/5.

Original entry on oeis.org

5, 10, 15, 20, 25, 30, 35, 40, 45, 50, 51, 52, 53, 54, 55, 60, 65, 70, 75, 80, 85, 90, 95, 100, 101, 102, 103, 104, 105, 110, 115, 120, 125, 130, 135, 140, 145, 150, 151, 152, 153, 154, 155, 160, 165, 170, 175, 180, 185, 190, 195, 200, 201, 202, 203, 204, 205, 210, 215, 220, 225
Offset: 1

Views

Author

Luca Onnis, Mar 20 2022

Keywords

Comments

All positive multiples of 5 are terms of the sequence.

Examples

			30 is a term, in fact A061078(30)=320, A061077(30)=400 and a(n) = 320/400 = is 4/5.
500, 501, 502, ..., 554, 555 are all terms. In fact 500=5*10^2 and for the formula above also 501, ..., 500+(5/9)*(10^2-1) = 555 are all terms of the sequence.

References

  • Amarnath Murthy, Smarandache friendly numbers and a few more sequences, Smarandache Notions Journal, Vol. 12, No. 1-2-3, Spring 2001.

Links

Crossrefs

Cf. A061076, A061077, A061078.

Programs

  • Mathematica
    Flatten[Position[(Accumulate[Times @@@ IntegerDigits[Range[2, 10000, 2]]]/
    Accumulate[Times @@@ IntegerDigits[Range[1, 9999, 2]]]), 4/5]]
  • PARI
    pd(n) = my(d = digits(n)); prod(i=1, #d, d[i]);
isok(k) = sum(i=1, k, pd(2*i))/sum(i=1, k, pd(2*i-1)) == 4/5; \\ Michel Marcus, Mar 21 2022

Formula

Let k be a positive integer not divisible by 5 and j >= 0; then 5*k*10^j, 5*k*10^j+1, ..., 5*k*10^j+(5/9)*(10^j-1) are all terms of the sequence.
Limit_{n->oo} A061078(n)/A061077(n) = 4/5.

A061077 a(n) is the sum of the products of the digits of the first n odd numbers.

Original entry on oeis.org

1, 4, 9, 16, 25, 26, 29, 34, 41, 50, 52, 58, 68, 82, 100, 103, 112, 127, 148, 175, 179, 191, 211, 239, 275, 280, 295, 320, 355, 400, 406, 424, 454, 496, 550, 557, 578, 613, 662, 725, 733, 757, 797, 853, 925, 934, 961, 1006, 1069, 1150, 1150, 1150, 1150
Offset: 1

Views

Author

Amarnath Murthy, Apr 14 2001

Keywords

Examples

			a(7) = 1 + 3 + 5 + 7 + 9 + 1*1 + 1*3 = 29.

References

  • Amarnath Murthy, Smarandache friendly numbers and a few more sequences, Smarandache Notions Journal, Vol. 12, No. 1-2-3, Spring 2001.

Links

Crossrefs

Cf. A061076, A061078.

Programs

  • Mathematica
    Accumulate[Times @@@ IntegerDigits[Range[1, 99, 2]]] (* Luca Onnis, Mar 20 2022 *)
  • PARI
    pd(n) = my(d = digits(n)); prod(i=1, #d, d[i]);
a(n) = sum(k=1, n, pd(2*k-1)); \\ Michel Marcus, Feb 01 2015
  • PARI
    a(n) = {m=digits(2*n - 1); p=1; d=#m; for(i=1, #m, if(m[i]==0, d=i-1; break));
(25/44) * (45^(#m-1)-1) + sum(i=1, #m-1, (prod(j=1,#m-i-1,m[j])) * (m[#m-i]) * (m[#m-i]-1) * (5^(i + 1) * 9^(i-1)) / 2)+prod(k=1,#m-1,m[k])*(((m[#m]+1)^2)/4)} \\ Luca Onnis, Mar 20 2022
  • Python
    from math import prod
def A061077(n): return sum(prod(int(d) for d in str(2*i+1)) for i in range(n)) # Chai Wah Wu, Mar 21 2022

Formula

a(n) = Sum_{k = 1..n} (product of the digits of 2k-1).
From Luca Onnis, Mar 20 2022: (Start)
a(5*10^n) = (25/44)*(45^(n+1)-1).
a(n) <= (25/44)*(45^(log(n/5)+1)-1) for all n.
a(n) ~ (5/4)*A061078(n) as n -> infinity. (End)

Extensions

More terms from Matthew Conroy, Apr 16 2001
Offset corrected by Charles R Greathouse IV, Feb 01 2015
Incorrect formula removed by Luca Onnis, Mar 20 2022

A360365 a(n) = sum of the products of the digits of the first n positive multiples of 3.

Original entry on oeis.org

3, 9, 18, 20, 25, 33, 35, 43, 57, 57, 66, 84, 111, 119, 139, 171, 176, 196, 231, 231, 249, 285, 339, 353, 388, 444, 452, 484, 540, 540, 567, 621, 702, 702, 702, 702, 703, 707, 714, 714, 720, 732, 750, 756, 771, 795, 799, 815, 843, 843, 858, 888, 933, 945, 975, 1023, 1030, 1058, 1107
Offset: 1

Views

Author

Luca Onnis, Feb 09 2023

Keywords

Examples

			a(4) = 20 since the first 4 positive multiples of 3 are 3,6,9 and 12 and the sum of the product of their digits is 3 + 6 + 9 + 1*2 = 20.

Crossrefs

Cf. A061076, A061077, A061078.

Programs

  • Mathematica
    Accumulate[Times @@@ IntegerDigits[Range[3, 999 , 3]]]
  • PARI
    a(n)={sum(k=1, n, vecprod(digits(3*k)))} \\ Andrew Howroyd, Feb 09 2023
  • Python
    from math import prod
def A360365(n): return sum(prod(int(d) for d in str(m)) for m in range(3,3*n+1,3)) # Chai Wah Wu, Feb 28 2023

Formula

a((10^n-1)/3) = (1/836)*(15*(19*45^n-63) + 44*3^((3*n)/2)*(sqrt(3)*sin((Pi*n)/6) + 15*cos((Pi*n)/6))).
G.f. of the subsequence a((10^n-1)/3): 9*(2 - 32*x + 135*x^2)/((1 - x)*(1 - 45*x)*(1 - 9*x + 27*x^2)).
a((10^n-1)/3) = 55*a((10^(n-1)-1)/3) - 486*a((10^(n-2)-1)/3) + 1647*a((10^(n-3)-1)/3) - 1215*a((10^(n-4)-1)/3) for n > 4.
Showing 1-4 of 4 results.

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