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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A350709 Modified Sisyphus function of order 3: a(n) is the concatenation of (number of digits of n)(number digits of n congruent to 0 modulo 3)(number of digits of n congruent to 1 modulo 3)(number of digits of n congruent to 2 modulo 3).

Original entry on oeis.org

1100, 1010, 1001, 1100, 1010, 1001, 1100, 1010, 1001, 1100, 2110, 2020, 2011, 2110, 2020, 2011, 2110, 2020, 2011, 2110, 2101, 2011, 2002, 2101, 2011, 2002, 2101, 2011, 2002, 2101, 2200, 2110, 2101, 2200, 2110, 2101, 2200
Offset: 0

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Author

Keywords

Comments

If we start with n and repeatedly apply the map i -> a(i), we eventually get the cycle {4031, 4112, 4220}

Examples

			11 has two digits, both congruent to 1 modulo 3, so a(11) = 2020.
a(20) = 2101.
a(30) = 2200.
a(1111123567) = 10262.
		

References

  • M. E. Coppenbarger, Iterations of a modified Sisyphus function, Fib. Q., 56 (No. 2, 2018), 130-141.

Crossrefs

Programs

  • Python
    def a(n):
        d, m = list(map(int, str(n))), [0, 0, 0]
        for di in d: m[di%3] += 1
        return int(str(len(d)) + "".join(map(str, m)))
    print([a(n) for n in range(37)]) # Michael S. Branicky, Mar 28 2022

A308005 A modified Sisyphus function: a(n) = concatenation of (number of odd digits in n) (number of digits in n) (number of even digits in n).

Original entry on oeis.org

11, 110, 11, 110, 11, 110, 11, 110, 11, 110, 121, 220, 121, 220, 121, 220, 121, 220, 121, 220, 22, 121, 22, 121, 22, 121, 22, 121, 22, 121, 121, 220, 121, 220, 121, 220, 121, 220, 121, 220, 22, 121, 22, 121, 22, 121, 22, 121, 22, 121, 121, 220, 121, 220, 121, 220, 121, 220, 121, 220, 22, 121, 22, 121, 22
Offset: 0

Views

Author

N. J. A. Sloane, May 12 2019

Keywords

Comments

If we start with n and repeatedly apply the map i -> a(i), it appears that we eventually reach one of the two fixed points 22 or 231, or enter the two-cycle (33, 220). Are there any other possibilities? This is in contrast to the behavior of the closely related A308003.

Examples

			11 has 2 digits, both odd, so a(11)=220.
12 has 2 digits, one even and one odd, so a(12)=121. Then a(121) = 231, a fixed point.
22 has two digits, both even, so 22 -> 22, another fixed point  (leading zeros are omitted).
		

References

  • M. E. Coppenbarger, Iterations of a modified Sisyphus function, Fib. Q., 56 (No. 2, 2018), 130-141.

Crossrefs

Programs

  • Maple
    Maple code based on R. J. Mathar's code for A171797:
    nevenDgs := proc(n) local a, d; a := 0 ; for d in convert(n, base, 10) do if type(d, 'even') then a :=a +1 ; end if; end do; a ; end proc:
    cat2 := proc(a, b) local ndigsb; ndigsb := max(ilog10(b)+1, 1) ; a*10^ndigsb+b ; end:
    catL := proc(L) local a, i; a := op(1, L) ; for i from 2 to nops(L) do a := cat2(a, op(i, L)) ; end do; a; end proc:
    A055642 := proc(n) max(1, ilog10(n)+1) ; end proc:
    A308005 := proc(n) local n1, n2 ; n1 := A055642(n) ; n2 := nevenDgs(n) ; catL([n1-n2, n1, n2]) ; end proc:
    [seq(A308005(n), n=0..80)];

A375208 Modified Sisyphus function of order 5.

Original entry on oeis.org

110000, 101000, 100100, 100010, 100001, 110000, 101000, 100100, 100010, 100001, 211000, 202000, 201100, 201010, 201001, 211000, 202000, 201100, 201010, 201001, 210100, 201100, 200200, 200110, 200101, 210100, 201100, 200200, 200110, 200101, 210010, 201010, 200110, 200020, 200011, 210010, 201010, 200110, 200020, 200011, 210001, 201001, 200101, 200011, 200002
Offset: 0

Views

Author

Matt Coppenbarger, Oct 16 2024

Keywords

Comments

a(n) is the concatenation of the number of digits in n with number of digits of n congruent to k modulo 5 for each k from 0 to 4 in turn. See Example.
If we start with n and repeatedly apply the map i -> a(i), we eventually get the cycle {613200, 622110}.

Examples

			11 has two digits, both congruent to 1 modulo 5, so a(11) = 202000.
a(20) = 210100.
a(30) = 210010.
a(2527200000) = 1060400.
		

Crossrefs

Programs

  • Maple
    a:= n-> (l-> parse(cat(nops(l), seq(add(`if`(irem(i, 5)=k
              , 1, 0), i=l), k=0..4))))(convert(n, base, 10)):
    seq(a(n), n=0..44);  # Alois P. Heinz, Oct 23 2024
  • Python
    # based on Michael S. Branicky in A350709
    def a(n, order=5):
        d, m = list(map(int, str(n))), [0]*order
        for di in d: m[di%order] += 1
        return int(str(len(d)) + "".join(map(str, m)))
    print([a(n) for n in range(37)])
    
  • Python
    from collections import Counter
    def A375208(n):
        s = str(n)
        c = Counter(int(d)%5 for d in s)
        return int(str(len(s))+''.join(str(c[i]) for i in range(5))) # Chai Wah Wu, Nov 26 2024

A108065 Numbers n such that DENEAT(n!) is prime, where DENEAT(n) = concatenate number of even digits in n, number of odd digits and total number of digits.

Original entry on oeis.org

1, 2, 3, 10, 15, 29, 35, 39, 51, 65, 167, 185, 198, 250, 282, 325, 366, 368, 382, 396, 400, 403, 450, 453, 509, 574, 575, 590, 598, 601, 699, 720, 759, 764, 788, 791, 797, 817, 824, 860, 863, 865, 867, 877, 901, 909, 911, 913, 930, 936, 1066, 1068, 1081, 1145
Offset: 1

Views

Author

Jason Earls, Jun 03 2005

Keywords

Examples

			10 is in the sequence because 10! = 3628800 has 6 even digits, 1
odd digit and 7 total digits, yielding the prime 617.
		

Crossrefs

Cf. A073053.

Programs

  • Mathematica
    eotQ[n_]:=Module[{idnf=IntegerDigits[n!],len,ev,od},len=Length[idnf];ev= Count[ idnf,?EvenQ];od=Count[idnf,?OddQ];PrimeQ[FromDigits[ Flatten[ IntegerDigits/@ Join[{ev,od,len}]]]]]; Select[Range[1200],eotQ] (* Harvey P. Dale, Jul 05 2017 *)

A108064 Numbers n such that DENEAT(n^n) is prime, where DENEAT(n) = concatenate number of even digits in n, number of odd digits and total number of digits.

Original entry on oeis.org

1, 2, 10, 12, 14, 26, 28, 34, 37, 44, 147, 156, 192, 229, 237, 246, 263, 282, 317, 325, 409, 413, 432, 436, 467, 510, 515, 534, 561, 570, 598, 600, 611, 636, 687, 702, 729, 738, 776, 818, 830, 859, 894, 901, 903, 914, 954, 1000, 1014, 1017, 1054, 1075, 1080
Offset: 1

Views

Author

Jason Earls, Jun 03 2005

Keywords

Examples

			12 is in the sequence because 12^12 = 8916100448256 has 9 even digits,
4 odd digits and 13 total digits, yielding the prime 9413.
		

Crossrefs

Cf. A073053.

Programs

  • Mathematica
    deneatQ[n_]:=Module[{idn=IntegerDigits[n^n]},PrimeQ[FromDigits[ Join[ IntegerDigits[ Count[ idn, ?EvenQ]],IntegerDigits[Count[idn,?OddQ]], IntegerDigits[Length[idn]]]]]]; Select[Range[1200],deneatQ] (* Harvey P. Dale, Aug 04 2015 *)

A352751 Modified Sisyphus function of order 4: a(n) is the concatenation of (number of digits of n)(number digits of n congruent to 0 modulo 4)(number of digits of n congruent to 1 modulo 4)(number of digits of n congruent to 2 modulo 4)(number of digits of n congruent to 3 modulo 4).

Original entry on oeis.org

11000, 10100, 10010, 10001, 11000, 10100, 10010, 10001, 11000, 10100, 21100, 20200, 20110, 20101, 21100, 20200, 20110, 20101, 21100, 20200, 21010, 20110, 20020, 20011, 21010, 20110, 20020, 20011, 21010, 20110, 21001, 20101, 20011, 20002, 21001, 20101, 20011, 20002, 21001, 20101, 22000, 21100, 21010
Offset: 0

Views

Author

Keywords

Comments

If we start with n and repeatedly apply the map i -> a(i), we eventually get one of three cycles: {51220}, {50410, 52111, 53200}, or {51301}

Examples

			11 has two digits, both congruent to 1 modulo 4, so a(11) = 20200.
a(20) = 21010.
a(30) = 21001.
a(1111123567) = 100622.
		

References

  • M. E. Coppenbarger, Iterations of a modified Sisyphus function, Fib. Q., 56 (No. 2, 2018), 130-141.

Crossrefs

Programs

  • Python
    def a(n, order=4):
        d, m = list(map(int, str(n))), [0]*order
        for di in d: m[di%order] += 1
        return int(str(len(d)) + "".join(map(str, m)))
    print([a(n) for n in range(37)]) # Michael S. Branicky, Apr 01 2022
Previous Showing 11-16 of 16 results.