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.

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A031007 Triangle T(n,k): Write n in base 7, reverse order of digits, to get row n.

Original entry on oeis.org

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

Views

Author

Clark Kimberling

Keywords

Crossrefs

Cf. A030308, A030341, A030386, A031235, A030567, A031045, A031087, A031298 for the base-2 to base-10 analogs.

Programs

  • Mathematica
    Flatten[Table[Reverse[IntegerDigits[n,7]],{n,0,50}]] (* Harvey P. Dale, Feb 25 2014 *)
  • PARI
    A031007(n, k=-1)={k<0&&error("Flattened sequence not yet implemented.");n\7^k%7} \\ Assuming that columns start with k=0 as in A030308, A030341, ... TO DO: implement flattened sequence, such that A030567(n)=a(n). - M. F. Hasler, Jul 21 2013

Extensions

Initial 0 and better name by Philippe Deléham, Oct 20 2011

A074847 Sum of 4-infinitary divisors of n: if n=Product p(i)^r(i) and d=Product p(i)^s(i), each s(i) has a digit a<=b in its 4-ary expansion everywhere that the corresponding r(i) has a digit b, then d is a 4-infinitary-divisor of n.

Original entry on oeis.org

1, 3, 4, 7, 6, 12, 8, 15, 13, 18, 12, 28, 14, 24, 24, 17, 18, 39, 20, 42, 32, 36, 24, 60, 31, 42, 40, 56, 30, 72, 32, 51, 48, 54, 48, 91, 38, 60, 56, 90, 42, 96, 44, 84, 78, 72, 48, 68, 57, 93, 72, 98, 54, 120, 72, 120, 80, 90, 60, 168, 62, 96, 104, 119, 84, 144, 68, 126, 96
Offset: 1

Views

Author

Yasutoshi Kohmoto, Sep 10 2002

Keywords

Comments

If we group the exponents e in the Bower-Harris formula into the sets with d_k=0, 1, 2 and 3, we see that every n has a unique representation of the form n=prod q_i *prod (r_j)^2 *prod (s_k)^3, where each of q_i, r_j, s_k is a prime power of the form p^(k^4), p prime, k>=0. Using this representation, a(n)=prod (q_i+1)prod ((r_j)^2+r_j+1)prod ((s_k)^3+(s_k)^2+s_k+1) by simple expansion of the quotient on the right hand side of the Bower-Harris formula. - Vladimir Shevelev, May 08 2013

Examples

			2^4*3 is a 4-infinitary-divisor of 2^5*3^2 because 2^4*3 = 2^10*3^1 and 2^5*3^2 = 2^11*3^2 in 4-ary expanded power. All corresponding digits satisfy the condition. 1<=1, 0<=1, 1<=2.

Links

Crossrefs

Cf. A049417 (2-infinitary), A049418 (3-infinitary), A097863 (5-infinitary).
Cf. A000302, A030386, A027748, A074848, A124010.

Programs

  • Haskell
    following Bower and Harris, cf. A049418:
a074847 1 = 1
a074847 n = product $ zipWith f (a027748_row n) (a124010_row n) where
   f p e = product $ zipWith div
           (map (subtract 1 . (p ^)) $
                zipWith (*) a000302_list $ map (+ 1) $ a030386_row e)
           (map (subtract 1 . (p ^)) a000302_list)
-- Reinhard Zumkeller, Sep 18 2015
  • Maple
    A074847 := proc(n) option remember; ifa := ifactors(n)[2] ; a := 1 ; if nops(ifa) = 1 then p := op(1,op(1,ifa)) ; e := op(2,op(1,ifa)) ; d := convert(e,base,4) ; for k from 0 to nops(d)-1 do a := a*(p^((1+op(k+1,d))*4^k)-1)/(p^(4^k)-1) ; end do: else for d in ifa do a := a*procname( op(1,d)^op(2,d)) ; end do: return a; end if; end proc:
seq(A074847(n),n=1..100) ; # R. J. Mathar, Oct 06 2010
  • Mathematica
    f[p_, e_] := Module[{d = IntegerDigits[e, 4]}, m = Length[d]; Product[(p^((d[[j]] + 1)*4^(m - j)) - 1)/(p^(4^(m - j)) - 1), {j, 1, m}]]; a[1] = 1; a[n_] := Times @@ f @@@ FactorInteger[n]; Array[a, 100] (* Amiram Eldar, Sep 09 2020 *)

Formula

Multiplicative. If e = sum_{k >= 0} d_k 4^k (base 4 representation), then a(p^e) = prod_{k >= 0} (p^(4^k*{d_k+1}) - 1)/(p^(4^k) - 1). - Christian G. Bower and Mitch Harris, May 20 2005

Extensions

More terms from R. J. Mathar, Oct 06 2010

A258059 Let n = Sum_{i=0..k} d_i*4^i be the base-4 expansion of n, with 0 <= d_i < 4. Then a(n) = minimal i such that d_i is not 1, or k+1 if there is no such i.

Original entry on oeis.org

1, 0, 0, 0, 2, 0, 0, 0, 1, 0, 0, 0, 1, 0, 0, 0, 1, 0, 0, 0, 3, 0, 0, 0, 1, 0, 0, 0, 1, 0, 0, 0, 1, 0, 0, 0, 2, 0, 0, 0, 1, 0, 0, 0, 1, 0, 0, 0, 1, 0, 0, 0, 2, 0, 0, 0, 1, 0, 0, 0, 1, 0, 0, 0, 1, 0, 0, 0, 2, 0, 0, 0, 1, 0, 0, 0, 1, 0, 0, 0, 1, 0, 0, 0, 4, 0, 0, 0, 1, 0, 0, 0, 1, 0, 0, 0, 1, 0, 0, 0, 2, 0, 0, 0, 1
Offset: 1

Views

Author

Richard C. Webster, May 17 2015

Keywords

Comments

This is the "General Ruler Sequence Base 4 Focused at 1" of Webster (2015).

Examples

			1 = 0*4+1, so a(1)=1.
7 = 1*4+3, so a(7)=0.
21 = 0*4^3+1*4^2+1*4+1, so a(21)=3.
523 base 10 is 20023 in base 4, so a(523)=0.
1365 base 10 is 111111 in base 4, so a(1365)=6.

Links

Crossrefs

The nonzero terms give A263845.
This sequence and A263845 are analogs of the pair of ruler sequences A007814 and A001511.
Cf. A007090, A235127.
Cf. A030386.

Programs

  • Haskell
    a258059 = f 0 . a030386_row where
   f i [] = i
   f i (t:ts) = if t == 1 then f (i + 1) ts else i
-- Reinhard Zumkeller, Nov 08 2015
  • Maple
    f:= proc(n)
  if n mod 4 = 1 then procname((n-1)/4) + 1 else 0 fi
end proc:
map(f, [$1..1000]); # Robert Israel, Jun 08 2015
  • PARI
    a(n) = {v = Vecrev(digits(n, 4)); for (i=1, #v, if (v[i] != 1, return (i-1));); return(#v);}

Formula

Recurrence: a(1)=1; thereafter a(4*n+1) = a(n)+1, a(4*n+j) = 0 for j = 0,2,3. G.f. g(x) = Sum_{k>=0} k * x^((4^k-1)/3) * (1 + x^(2*4^k) + x^(3*4^k))/(1 - x^(4*4^k)) satisfies g(x) = x*g(x^4) + x/(1-x^4). - Robert Israel, Jun 08 2015

Extensions

Edited by N. J. A. Sloane, Oct 31 2015 and Nov 06 2015.
Previous Showing 21-23 of 23 results.

Started in 1964 by Neil J. A. Sloane | Maintained by The OEIS Foundation Inc.

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