A384877 -id:A384877 - cp's OEIS frontend

A385886 Irregular triangle read by rows listing the lengths of maximal anti-runs (sequences of distinct consecutive elements increasing by more than 1) of binary indices, duplicate rows removed.

1, 1, 1, 2, 1, 1, 1, 1, 2, 2, 1, 1, 1, 1, 1, 3, 1, 1, 2, 1, 2, 1, 2, 1, 1, 1, 1, 1, 1, 1, 1, 3, 2, 2, 1, 1, 1, 2, 3, 1, 1, 1, 2, 1, 1, 2, 1, 1, 2, 1, 1, 1, 1, 1, 1, 1, 1, 1, 4, 1, 1, 3, 1, 2, 2, 2, 1, 2, 1, 1, 1, 1, 2, 1, 3, 1, 2, 2, 1, 1, 1, 1, 2, 1

Offset: 0

Gus Wiseman, Jul 14 2025

A binary index of n is any position of a 1 in its reversed binary expansion. The binary indices of n are row n of A048793.

This is the triangle A384877, except all duplicates after the first instance of each composition are removed. It lists all compositions in order of their first appearance as a row of A384877.

			The binary indices of 27 are {1,2,4,5}, with maximal anti-runs ((1),(2,4),(5)), with lengths (1,2,1). After removing duplicates, this is our row 10.
The binary indices of 53 are {1,3,5,6}, with maximal anti-runs ((1,3,5),(6)), with lengths (3,1). After removing duplicates, this is our row 16.
Triangle begins:
   0: .
   1: 1
   2: 1 1
   3: 2
   4: 1 1 1
   5: 1 2
   6: 2 1
   7: 1 1 1 1
   8: 3
   9: 1 1 2
  10: 1 2 1
  11: 2 1 1
  12: 1 1 1 1 1
  13: 1 3
  14: 2 2
  15: 1 1 1 2
  16: 3 1
  17: 1 1 2 1
  18: 1 2 1 1
  19: 2 1 1 1
  20: 1 1 1 1 1 1

In the following references, "before" is short for "before removing duplicate rows".
Positions of singleton rows appear to be A001906 = A055588 - 1.
Positions of rows of the form (1,1,...) appear to be A001911-2, before A023758.
Row sums appear to be A200648, before A000120.
Row lengths appear to be A200649, before A384890.
Standard composition numbers of each row appear to be A348366.
Before we had A384877, ranks A385816, firsts A052499.
For runs instead of anti-runs we have A385817, see A245563, A245562, A246029.
Cf. A044813, A048793, A069010, A164707, A242882, A328592, A384175, A384177, A384878, A384879, A384893.

DeleteDuplicates[Table[Length/@Split[Join@@Position[Reverse[IntegerDigits[n,2]],1],#2!=#1+1&],{n,0,100}]]

A209859 Rewrite the binary expansion of n from the most significant end, 1 -> 1, 0+1 (one or more zeros followed by one) -> 0, drop the trailing zeros of the original n.

Original entry on oeis.org

0, 1, 1, 3, 1, 2, 3, 7, 1, 2, 2, 5, 3, 6, 7, 15, 1, 2, 2, 5, 2, 4, 5, 11, 3, 6, 6, 13, 7, 14, 15, 31, 1, 2, 2, 5, 2, 4, 5, 11, 2, 4, 4, 9, 5, 10, 11, 23, 3, 6, 6, 13, 6, 12, 13, 27, 7, 14, 14, 29, 15, 30, 31, 63, 1, 2, 2, 5, 2, 4, 5, 11, 2, 4, 4, 9, 5, 10, 11, 23, 2, 4, 4, 9, 4, 8, 9, 19, 5, 10, 10, 21, 11, 22, 23, 47, 3, 6, 6, 13, 6, 12, 13, 27, 6, 12, 12, 25, 13

Offset: 0

Antti Karttunen, Mar 24 2012

This is the number k such that the k-th composition in standard order is the reversed sequence of lengths of the maximal anti-runs of the binary indices of n. Here, the binary indices of n are row n of A048793, and the k-th composition in standard order is row k of A066099. For example, the binary indices of 100 are {3,6,7}, with maximal anti-runs ((3,6),(7)), with reversed lengths (1,2), which is the 6th composition in standard order, so a(100) = 6. - Gus Wiseman, Jul 27 2025

			102 in binary is 1100110, we rewrite it from the left so that first two 1's stay same ("11"), then "001" is rewritten to "0", the last 1 to "1", and we ignore the last 0, thus getting 1101, which is binary expansion of 13, thus a(102) = 13.

Gus Wiseman, Statistics, classes, and transformations of standard compositions

This is an "inverse" of A071162, i.e. a(A071162(n)) = n for all n. Bisection: A209639. Used to construct permutation A209862.
Removing duplicates appears to give A358654.
Sorted positions of firsts appearances appear to be A247648+1.
A245563 lists run-lengths of binary indices (ranks A246029), reverse A245562.
A384175 counts subsets with all distinct lengths of maximal runs, complement A384176.
Cf. A000120, A023758, A029931, A044813, A048793, A052499, A069010, A348366, A384878, A384879, A384893.
Cf. A384877, A385816, A385887, A385889.

bpe[n_]:=Join@@Position[Reverse[IntegerDigits[n,2]],1];
stcinv[q_]:=Total[2^(Accumulate[Reverse[q]])]/2;
Table[stcinv[Reverse[Length/@Split[bpe[n],#2!=#1+1&]]],{n,0,100}] (* Gus Wiseman, Jul 25 2025 *)

import re
def a(n): return int(re.sub("0+1", "0", bin(n)[2:].rstrip("0")), 2) if n else 0
print([a(n) for n in range(109)])  # Michael S. Branicky, Jul 25 2025

(define (A209859 n) (let loop ((n n) (s 0) (i (A053644 n))) (cond ((zero? n) s) ((> i n) (if (> (/ i 2) n) (loop n s (/ i 2)) (loop (- n (/ i 2)) (* 2 s) (/ i 4)))) (else (loop (- n i) (+ (* 2 s) 1) (/ i 2))))))

a(n) = a(A000265(n)).

A385889 The number k such that the k-th composition in standard order is the sequence of lengths of maximal runs of binary indices of n.

Original entry on oeis.org

0, 1, 1, 2, 1, 3, 2, 4, 1, 3, 3, 5, 2, 6, 4, 8, 1, 3, 3, 5, 3, 7, 5, 9, 2, 6, 6, 10, 4, 12, 8, 16, 1, 3, 3, 5, 3, 7, 5, 9, 3, 7, 7, 11, 5, 13, 9, 17, 2, 6, 6, 10, 6, 14, 10, 18, 4, 12, 12, 20, 8, 24, 16, 32, 1, 3, 3, 5, 3, 7, 5, 9, 3, 7, 7, 11, 5, 13, 9, 17, 3

Offset: 0

Gus Wiseman, Jul 16 2025

nonn

A binary index of n is any position of a 1 in its reversed binary expansion. The binary indices of n are row n of A048793.

The k-th composition in standard order (graded reverse-lexicographic, A066099) is obtained by taking the set of positions of 1's in the reversed binary expansion of k, prepending 0, taking first differences, and reversing again. This gives a bijective correspondence between nonnegative integers and integer compositions.

			The binary indices of 27 are {1,2,4,5}, with maximal runs ((1,2),(4,5)), with lengths (2,2), which is the 10th composition in standard order, so a(27) = 10.
The binary indices of 100 are {3,6,7}, with maximal runs ((3),(6,7)), with lengths (1,2), which is the 6th composition in standard order, so a(100) = 6.

Gus Wiseman, Statistics, classes, and transformations of standard compositions

Sorted positions of firsts appearances appear to be A247648+1.
After removing duplicates we get A385818.
The reverse version is A385887.
A245563 lists run lengths of binary indices (ranks A246029), reverse A245562.
A384877 lists anti-run lengths of binary indices (ranks A385816), reverse A209859.
Cf. A000120, A023758, A029931, A044813, A048793, A069010, A348366, A384175, A384878, A385817.

bpe[n_]:=Join@@Position[Reverse[IntegerDigits[n,2]],1];
stcinv[q_]:=Total[2^(Accumulate[Reverse[q]])]/2;
Table[stcinv[Length/@Split[bpe[n],#2==#1+1&]],{n,0,100}]

A245564 a(n) = Product_{i in row n of A245562} Fibonacci(i+2).

Original entry on oeis.org

1, 2, 2, 3, 2, 4, 3, 5, 2, 4, 4, 6, 3, 6, 5, 8, 2, 4, 4, 6, 4, 8, 6, 10, 3, 6, 6, 9, 5, 10, 8, 13, 2, 4, 4, 6, 4, 8, 6, 10, 4, 8, 8, 12, 6, 12, 10, 16, 3, 6, 6, 9, 6, 12, 9, 15, 5, 10, 10, 15, 8, 16, 13, 21, 2, 4, 4, 6, 4, 8, 6, 10, 4, 8, 8, 12, 6, 12, 10, 16, 4, 8, 8, 12, 8, 16, 12, 20, 6, 12, 12, 18

Offset: 0

N. J. A. Sloane, Aug 10 2014; revised Sep 05 2014

This is the Run Length Transform of S(n) = Fibonacci(n+2).
The Run Length Transform of a sequence {S(n), n>=0} is defined to be the sequence {T(n), n>=0} given by T(n) = Product_i S(i), where i runs through the lengths of runs of 1's in the binary expansion of n. E.g. 19 is 10011 in binary, which has two runs of 1's, of lengths 1 and 2. So T(19) = S(1)*S(2). T(0)=1 (the empty product).
Also the number of sparse subsets of the binary indices of n, where a set is sparse iff 1 is not a first difference. The maximal case is A384883. For prime instead of binary indices we have A166469. - Gus Wiseman, Jul 05 2025

			From _Gus Wiseman_, Jul 05 2025: (Start)
The binary indices of 11 are {1,2,4}, with sparse subsets {{},{1},{2},{4},{1,4},{2,4}}, so a(11) = 6.
The maximal runs of binary indices of 11 are ((1,2),(4)), with lengths (2,1), so a(11) = F(2+2)*F(1+2) = 6.
The a(0) = 1 through a(12) = 3 sparse subsets are:
  0    1    2    3    4    5    6    7    8    9    10    11    12
  ------------------------------------------------------------------
  {}   {}   {}   {}   {}   {}   {}   {}   {}   {}    {}    {}    {}
       {1}  {2}  {1}  {3}  {1}  {2}  {1}  {4}  {1}   {2}   {1}   {3}
                 {2}       {3}  {3}  {2}       {4}   {4}   {2}   {4}
                           {1,3}     {3}       {1,4} {2,4} {4}
                                     {1,3}                 {1,4}
                                                           {2,4}
The greatest number whose set of binary indices is a member of column n above is A374356(n).
(End)

Alois P. Heinz, Table of n, a(n) for n = 0..8191
Chai Wah Wu, Sums of products of binomial coefficients mod 2 and run length transforms of sequences, arXiv:1610.06166 [math.CO], 2016.

Cf. A245562, A000045, A001045, A071053, A245565, A246028.
A034839 counts subsets by number of maximal runs, strict partitions A116674.
A384877 gives lengths of maximal anti-runs of binary indices, firsts A384878.
A384893 counts subsets by number of maximal anti-runs, for partitions A268193, A384905.
Cf. A000071, A001629, A010049, A053538, A166469, A202023, A202064, A208342, A374356, A384883.

with(combinat); ans:=[];
for n from 0 to 100 do lis:=[]; t1:=convert(n,base,2); L1:=nops(t1); out1:=1; c:=0;
for i from 1 to L1 do
   if out1 = 1 and t1[i] = 1 then out1:=0; c:=c+1;
   elif out1 = 0 and t1[i] = 1 then c:=c+1;
   elif out1 = 1 and t1[i] = 0 then c:=c;
   elif out1 = 0 and t1[i] = 0 then lis:=[c,op(lis)]; out1:=1; c:=0;
   fi;
   if i = L1 and c>0 then lis:=[c,op(lis)]; fi;
                   od:
a:=mul(fibonacci(i+2), i in lis);
ans:=[op(ans),a];
od:
ans;

a[n_] := Sum[Mod[Binomial[3k, k] Binomial[n, k], 2], {k, 0, n}];
a /@ Range[0, 100] (* Jean-François Alcover, Feb 29 2020, after Chai Wah Wu *)
spars[S_]:=Select[Subsets[S],FreeQ[Differences[#],1]&];
bpe[n_]:=Join@@Position[Reverse[IntegerDigits[n,2]],1];
Table[Length[spars[bpe[n]]],{n,0,30}] (* Gus Wiseman, Jul 05 2025 *)

a(n)=my(s=1,k); while(n, n>>=valuation(n,2); k=valuation(n+1,2); s*=fibonacci(k+2); n>>=k); s \\ Charles R Greathouse IV, Oct 21 2016

# use RLT function from A278159
from sympy import fibonacci
def A245564(n): return RLT(n,lambda m: fibonacci(m+2)) # Chai Wah Wu, Feb 04 2022

a(n) = Sum_{k=0..n} ({binomial(3k,k)*binomial(n,k)} mod 2). - Chai Wah Wu, Oct 19 2016

A385817 Irregular triangle read by rows listing the lengths of maximal runs (sequences of consecutive elements increasing by 1) of binary indices, duplicate rows removed.

Original entry on oeis.org

1, 2, 1, 1, 3, 2, 1, 1, 2, 4, 1, 1, 1, 3, 1, 2, 2, 1, 3, 5, 2, 1, 1, 1, 2, 1, 4, 1, 1, 1, 2, 3, 2, 2, 3, 1, 4, 6, 1, 1, 1, 1, 3, 1, 1, 2, 2, 1, 1, 3, 1, 5, 1, 2, 1, 2, 1, 2, 2, 4, 2, 1, 1, 3, 3, 3, 2, 4, 1, 5, 7, 2, 1, 1, 1, 1, 2, 1, 1, 4, 1, 1, 1, 1, 2, 1

Offset: 0

Gus Wiseman, Jul 14 2025

A binary index of n is any position of a 1 in its reversed binary expansion. The binary indices of n are row n of A048793.

This is the triangle A245563, except all duplicates after the first instance of each composition are removed. It lists all compositions in order of their first appearance as a row of A245563.

			The binary indices of 53 are {1,3,5,6}, with maximal runs ((1),(3),(5,6)), with lengths (1,1,2). After removing duplicates, this is our row 16.
Triangle begins:
   0: .
   1: 1
   2: 2
   3: 1 1
   4: 3
   5: 2 1
   6: 1 2
   7: 4
   8: 1 1 1
   9: 3 1
  10: 2 2
  11: 1 3
  12: 5
  13: 2 1 1
  14: 1 2 1
  15: 4 1
  16: 1 1 2
  17: 3 2
  18: 2 3
  19: 1 4
  20: 6
  21: 1 1 1 1

In the following references, "before" is short for "before removing duplicate rows".
Positions of singleton rows appear to be A000071 = A000045-1, before A023758.
Positions of firsts appearances appear to be A001629.
Positions of rows of the form (1,1,...) appear to be A055588 = A001906+1.
First term of each row appears to be A083368.
Row sums appear to be A200648, before A000120.
Row lengths after the first row appear to be A200650+1, before A069010 = A037800+1.
Before the removals we had A245563 (except first term), see A245562, A246029, A328592.
For anti-run ranks we have A385816, before A348366, firsts A052499.
Standard composition numbers of rows are A385818, before A385889.
For anti-runs we have A385886, before A384877, firsts A384878.
Cf. A044813, A048793, A164707, A200649, A242882, A243815, A384175, A384890.

DeleteDuplicates[Table[Length/@Split[Join@@Position[Reverse[IntegerDigits[n,2]],1],#2==#1+1&],{n,0,100}]]

A083368 a(n) is the position of the highest one in A003754(n).

Original entry on oeis.org

1, 2, 1, 3, 2, 1, 4, 1, 3, 2, 1, 5, 2, 1, 4, 1, 3, 2, 1, 6, 1, 3, 2, 1, 5, 2, 1, 4, 1, 3, 2, 1, 7, 2, 1, 4, 1, 3, 2, 1, 6, 1, 3, 2, 1, 5, 2, 1, 4, 1, 3, 2, 1, 8, 1, 3, 2, 1, 5, 2, 1, 4, 1, 3, 2, 1, 7, 2, 1, 4, 1, 3, 2, 1, 6, 1, 3, 2, 1, 5, 2, 1, 4, 1, 3, 2, 1, 9, 2, 1, 4, 1, 3, 2, 1, 6, 1, 3, 2, 1, 5, 2

Offset: 1

Gary W. Adamson, Jun 04 2003

Previous name was: A Fibbinary system represents a number as a sum of distinct Fibonacci numbers (instead of distinct powers of two). Using representations without adjacent zeros, a(n) = the highest bit-position which changes going from n-1 to n.
A003754(n), when written in binary, is the representation of n.
Often one uses Fibbinary representations without adjacent ones (the Zeckendorf expansion).
a(A000071(n+3)) = n. - Reinhard Zumkeller, Aug 10 2014
From Gus Wiseman, Jul 24 2025: (Start)
Conjecture: To obtain this sequence, start with A245563 (maximal run lengths of binary indices), then remove empty and duplicate rows (giving A385817), then take the first term of each remaining row. Some variations:
- For sum instead of first term we appear to have A200648.
- For length instead of first term we appear to have A200650+1.
- For last instead of first term we have A385892.
(End)

			27 is represented 110111, 28 is 111010; the fourth position changes, so a(28)=4.

Jay Kappraff, Beyond Measure: A Guided Tour Through Nature, Myth and Number, World Scientific, 2002, page 460.

Reinhard Zumkeller, Table of n, a(n) for n = 1..10000

A035612 is the analogous sequence for Zeckendorf representations.
A001511 is the analogous sequence for power-of-two representations.
Cf. A003714, A003754.
Cf. A000045, A000071.
Appears to be the first element of each row of A385817, see A083368, A200648, A200650, A385818, A385892.
A000120 counts ones in binary expansion.
A245563 gives run lengths of binary indices, see A089309, A090996, A245562, A246029, A328592.
A384877 gives anti-run lengths of binary indices, ranks A385816.
Cf. A001629, A037800, A044813, A048793, A069010, A200649, A384878, A385886.

a083368 n = a083368_list !! (n-1)
a083368_list = concat $ h $ drop 2 a000071_list where
   h (a:fs@(a':_)) = (map (a035612 . (a' -)) [a .. a' - 1]) : h fs
-- Reinhard Zumkeller, Aug 10 2014

For n = F(a)-1 to F(a+1)-2, a(n) = A035612(F(a+1)-1-n).

a(n) = a(k)+1 if n = ceiling(phi*k) where phi is the golden ratio; otherwise a(n) = 1. - Tom Edgar, Aug 25 2015

Edited by Don Reble, Nov 12 2005

Shorter name from Joerg Arndt, Jul 27 2025

A385818 The number k such that the k-th composition in standard order lists the maximal run lengths of each nonnegative integer's binary indices, with duplicates removed.

Original entry on oeis.org

0, 1, 2, 3, 4, 5, 6, 8, 7, 9, 10, 12, 16, 11, 13, 17, 14, 18, 20, 24, 32, 15, 19, 21, 25, 33, 22, 26, 34, 28, 36, 40, 48, 64, 23, 27, 35, 29, 37, 41, 49, 65, 30, 38, 42, 50, 66, 44, 52, 68, 56, 72, 80, 96, 128, 31, 39, 43, 51, 67, 45, 53, 69, 57, 73, 81, 97

Offset: 0

Gus Wiseman, Jul 18 2025

A permutation of the nonnegative integers.
A binary index of n is any position of a 1 in its reversed binary expansion. The binary indices of n are row n of A048793.
The k-th composition in standard order (graded reverse-lexicographic, A066099) is obtained by taking the set of positions of 1's in the reversed binary expansion of k, prepending 0, taking first differences, and reversing again. This gives a bijective correspondence between nonnegative integers and integer compositions.

			The binary indices of 53 are {1,3,5,6}, with maximal runs ((1),(3),(5,6)) with lengths (1,1,2), which is the 14th composition in standard order, so A385889(53) = 14, and after removing duplicate rows a(16) = 14.

John Tyler Rascoe, Table of n, a(n) for n = 0..8192

For anti-runs instead of runs we appear to have A348366.
See also A385816 (standard compositions of rows of A384877), reverse A209859.
The compositions themselves are listed by A385817.
Before removing duplicates we had A385889.
A245563 lists run lengths of binary indices (ranks A246029), rev A245562, strict A328592.
A384175 counts subsets with all distinct lengths of maximal runs, complement A384176.
Cf. A000120, A029931, A048793, A052499, A069010, A200648, A200649, A200650, A384890, A385886, A385887.

bpe[n_]:=Join@@Position[Reverse[IntegerDigits[n,2]],1];
stcinv[q_]:=Total[2^(Accumulate[Reverse[q]])]/2;
stcinv/@DeleteDuplicates[Table[Length/@Split[bpe[n],#2==#1+1&],{n,0,100}]]

A384883 Number of maximal sparse subsets of the binary indices of n, where a set is sparse iff 1 is not a first difference.

Original entry on oeis.org

1, 1, 1, 2, 1, 1, 2, 2, 1, 1, 1, 2, 2, 2, 2, 3, 1, 1, 1, 2, 1, 1, 2, 2, 2, 2, 2, 4, 2, 2, 3, 4, 1, 1, 1, 2, 1, 1, 2, 2, 1, 1, 1, 2, 2, 2, 2, 3, 2, 2, 2, 4, 2, 2, 4, 4, 2, 2, 2, 4, 3, 3, 4, 5, 1, 1, 1, 2, 1, 1, 2, 2, 1, 1, 1, 2, 2, 2, 2, 3, 1, 1, 1, 2, 1, 1, 2

Offset: 0

Gus Wiseman, Jul 02 2025

nonn

A binary index of n is any position of a 1 in its reversed binary expansion. The binary indices of n are row n of A048793.

			The binary indices of 27 are {1,2,4,5}, with maximal sparse subsets {{1,4},{1,5},{2,4},{2,5}}, so a(27) = 4.

For subsets of {1..n} we get A000931 (shifted), maximal case of A000045 (shifted).
This is the maximal case of A245564.
The greatest number whose binary indices are one of these subsets is A374356.
For prime instead of binary indices we have A385215, maximal case of A166469.
A034839 counts subsets by number of maximal runs, for strict partitions A116674.
A202064 counts subsets containing n with k maximal runs.
A384877 gives lengths of maximal anti-runs in binary indices, firsts A384878.
A384893 counts subsets by number of maximal anti-runs, for partitions A268193, A384905.
Cf. A000071, A001629, A010049, A044813, A053538, A119900, A202023, A208342, A384177, A384890.

spars[S_]:=Select[Subsets[S],FreeQ[Differences[#],1]&];
bpe[n_]:=Join@@Position[Reverse[IntegerDigits[n,2]],1];
maximize[sys_]:=Complement@@Prepend[Most[Subsets[#]]&/@sys,sys];
Table[Length[maximize[spars[bpe[n]]]],{n,0,100}]

A384906 Number of maximal anti-runs of consecutive parts not increasing by 1 in the prime indices of n (with multiplicity).

Original entry on oeis.org

0, 1, 1, 1, 1, 2, 1, 1, 1, 1, 1, 2, 1, 1, 2, 1, 1, 2, 1, 1, 1, 1, 1, 2, 1, 1, 1, 1, 1, 3, 1, 1, 1, 1, 2, 2, 1, 1, 1, 1, 1, 2, 1, 1, 2, 1, 1, 2, 1, 1, 1, 1, 1, 2, 1, 1, 1, 1, 1, 3, 1, 1, 1, 1, 1, 2, 1, 1, 1, 2, 1, 2, 1, 1, 2, 1, 2, 2, 1, 1, 1, 1, 1, 2, 1, 1, 1

Offset: 1

Gus Wiseman, Jun 22 2025

nonn

First differs from A300820 at a(462) = 3, A300820(462) = 2.

A prime index of n is a number m such that prime(m) divides n. The multiset of prime indices of n is row n of A112798.

			The prime indices of 462 are {1,2,4,5}, with maximal anti-runs ((1),(2,4),(5)), so a(462) = 3.

For the strict case we have A356228.
For binary instead of prime indices we have A384890 (for runs A069010).
For runs instead of anti-runs we have A385213.
A034839 counts subsets by number of maximal runs, for strict partitions A116674.
A055396 gives least prime index, greatest A061395.
A056239 adds up prime indices, row sums of A112798.
A384877 gives lengths of maximal anti-runs in binary indices, firsts A384878.
Cf. A130091, A245562, A268193, A300820, A351202, A356607, A382525, A384177, A384321, A384893.

prix[n_]:=If[n==1,{},Flatten[Cases[FactorInteger[n],{p_,k_}:>Table[PrimePi[p],{k}]]]];
Table[Length[Split[prix[n],#2!=#1+1&]],{n,100}]

A385213 Number of maximal runs of consecutive parts increasing by 1 in the prime indices of n (with multiplicity).

Original entry on oeis.org

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

Offset: 1

Gus Wiseman, Jun 22 2025

nonn

A prime index of n is a number m such that prime(m) divides n. The multiset of prime indices of n is row n of A112798.

			The prime indices of 24 are {1,1,1,2}, with maximal runs ((1),(1),(1,2)), so a(24) = 3.

Positions of first appearances are A000079.
For binary instead of prime indices we have A069010 (for anti-runs A384890).
For anti-runs instead of runs we have A384906.
A034839 counts subsets by number of maximal runs, for strict partitions A116674.
A055396 gives least prime index, greatest A061395.
A056239 adds up prime indices, row sums of A112798.
A384877 gives lengths of maximal anti-runs in binary indices, firsts A384878.
Cf. A002110, A048767, A130091, A300820, A356606, A351202, A382525, A384893.

prix[n_]:=If[n==1,{},Flatten[Cases[FactorInteger[n],{p_,k_}:>Table[PrimePi[p],{k}]]]];
Table[Length[Split[prix[n],#2==#1+1&]],{n,100}]

cp's OEIS Frontend

A385886 Irregular triangle read by rows listing the lengths of maximal anti-runs (sequences of distinct consecutive elements increasing by more than 1) of binary indices, duplicate rows removed.

Views

Author

Keywords

Comments

Examples

Crossrefs

Programs

Mathematica

A209859 Rewrite the binary expansion of n from the most significant end, 1 -> 1, 0+1 (one or more zeros followed by one) -> 0, drop the trailing zeros of the original n.

Views

Author

Keywords

Comments

Examples

Links

Crossrefs

Programs

Mathematica

Python

Scheme

Formula

A385889 The number k such that the k-th composition in standard order is the sequence of lengths of maximal runs of binary indices of n.

Views

Author

Keywords

Comments

Examples

Links

Crossrefs

Programs

Mathematica

A245564 a(n) = Product_{i in row n of A245562} Fibonacci(i+2).

Views

Author

Keywords

Comments

Examples

Links

Crossrefs

Programs

Maple

Mathematica

PARI

Python

Formula

A385817 Irregular triangle read by rows listing the lengths of maximal runs (sequences of consecutive elements increasing by 1) of binary indices, duplicate rows removed.

Views

Author

Keywords

Comments

Examples

Crossrefs

Programs

Mathematica

A083368 a(n) is the position of the highest one in A003754(n).

Views

Author

Keywords

Comments

Examples

References

Links

Crossrefs

Programs

Haskell

Formula

Extensions

A385818 The number k such that the k-th composition in standard order lists the maximal run lengths of each nonnegative integer's binary indices, with duplicates removed.

Views

Author

Keywords

Comments

Examples

Links

Crossrefs

Programs

Mathematica

A384883 Number of maximal sparse subsets of the binary indices of n, where a set is sparse iff 1 is not a first difference.