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.

A221741 a(n) = -4*(((n+1)^(n+1)-(n+1))/((n+1)-1)^2-1)/((-3+(-1)^n)*n).

Original entry on oeis.org

1, 5, 9, 97, 373, 7625, 48913, 1513361, 13717421, 570623341, 6698798233, 350549891889, 5057809205989, 319164643134737, 5465701947765793, 403925909124187873, 8008631808527689309, 678470389458269406421, 15287592943577781017641
Offset: 1

Views

Author

Alexander R. Povolotsky, Jan 23 2013

Keywords

Comments

Per exhaustive program, written for bases from 2 to 10, the number of permutations pairs, which have the same ratio, equal to A221740(n)/a(n) = (n^2 (n+1)^n-(n+1)^n+1) / (-n^2+n (n+1)^n+(n+1)^n-n-1), is: {2,2,3,3,5,3,7,5,7,...} for n>=1 where n=r-1 and r is the base radix. Judging by above sequence it appears that the number of such permutations pairs is related to phi, which is the Euler totient function - according to A039649, A039650, A214288 (see bullet 1 of the analysis in the answer section of the StackExchange link). Alexander R. Povolotsky, Jan 26 2013

Links

Crossrefs

Cf. A221740, A023811.

Programs

  • Mathematica
    Table[-4*(((n + 1)^(n + 1) - (n + 1))/((n + 1) - 1)^2 - 1)/((-3 + (-1)^n)*n), {n,1,50}] (* G. C. Greubel, Feb 19 2017 *)
  • Maxima
    makelist(-4*(((n+1)^(n+1)-(n+1))/((n+1)-1)^2-1)/((-3+(-1)^n)*n), n, 1, 20); /* Martin Ettl, Jan 25 2013 */
  • PARI
    for(n=1,25, print1(-4*(((n + 1)^(n + 1) - (n + 1))/((n + 1) - 1)^2 - 1)/((-3 + (-1)^n)*n), ", ")) \\ G. C. Greubel, Feb 19 2017

Formula

a(n) = -4*A023811(n+1)/((-3 + (-1)^n)*n).

A051847 Bisection of A051846, divided by the term position.

Original entry on oeis.org

1, 19, 1493, 293479, 109739369, 66987982331, 60693710471869, 76519827268721103, 128138108936443028945, 275176672984400058317539, 737345594135016860806925221, 2411620538399461719230688945719
Offset: 1

Views

Author

Antti Karttunen, Dec 13 1999

Keywords

Crossrefs

Cf. A051846, A221740.

Formula

a(n) = A051846(2n-1)/(2n-1).
a(n) = A221740(2n-1). - Alexander R. Povolotsky, Oct 13 2022

A212958 Array with a variable number of columns, where terms in the n-th row are the differences (computed in decimal base and divided by 9) between equal ratio permutations, found in the base n>=2, and the first (in ascending order of digits) minimal value permutation of {0,1,...,n}.

Original entry on oeis.org

0, 0, 1, 1, 0, 1, 12, 22, 0, 21, 22, 123, 131, 343, 0, 342, 343, 1234, 2531, 4664, 0, 1421, 3242, 4663, 12345, 58985, 0, 58984, 58985, 23456, 497531, 713306, 0, 137421, 276842, 436463, 575884, 713305, 713306, 1234567, 1810675, 2907844, 4002993, 6197531, 8367727
Offset: 1

Views

Author

Alexander R. Povolotsky, Feb 14 2013

Keywords

Comments

It is conjectured (conjecture #1 - based on the observation of the programmatical exhaustive computation results) that among all possible distinct {0,1,...,n} permutations for each base n>=2, there are at least two pairs of such permutations that yield the same ratio, equal to A221740(p)/A221741(p) = (p^2*(p+1)^p -(p+1)^p+1)/(-p^2+p*(p+1)^p+(p+1)^p-p-1) where p=n-1. Additionally it is conjectured (conjecture #2) that the number of such pairs for each n is equal to A039649(p).
Construction of the terms for the n-th row of this sequence (where n is the base, n>=2) requires pre-calculation of the (mentioned in the above conjecture #1) permutation pairs (see the StackExchange link) that have the same ratio. Given the known (as defined above) ratio, actual values of the permutation pairs for each n (there are A039649(p) such pairs for each n>=2 - as conjectured above) are findable either by trial and error, or via executing an exhaustive computer program (see the link, featuring PARI/GP program - detection script for finding the equal-ratio permutation pairs for each base in the range from 2 to 10).
The terms of this sequence should be viewed in the table layout. The above mentioned table is an array with a variable number of columns. The number of terms (columns)in n-th row is, as conjectured above, equal to 2*A039649(p). Each n-th row (rows are counted from n=2) starts with a zero term because (as conjectured - conjecture #3) the 1st in ascending order minimal value permutation of {0,1,...,n} is always present among elements of the pairs, and, as it appears, always ends with the a(n=sum(i=2...n,2*A039649(i))) term, which has a nonzero value, and always seems to be equal to A215940((n)!) - conjecture #4.
Denoting the decimal value of the number made by the concatenation of the digits of the first (identical) base-n permutation of {0,1,...,n} as Pn, and considering terms of this sequence a(n) as members of a two-dimensional array b(n,k), where n is the row number and k is the position number in the n-th row (n>1, 1 <= k <=2*A039649(p)), then (per conjecture #1) it is true that in each array row "n" there are A039649(p) pairs of decimal integer values j and l such that (b(n,j)+ Pn)/(b(n,l)+ Pn) = A221740(p)/A221741(p) for n>=2, 1 <= j <= A039649(p), 1 <= l <= j.
It also appears that in the "next" (n+1)-th row there is always present a term that is less by one in decimal value as compared with the ending term in the "previous" n-th row (conjecture #5).

Examples

			For the fourth (n=4) row, which relates to base-4 four-digit {0, 1, 2, 3} distinct permutations, there are A039649(p) pairs where p = n-1 and thus for n=4, p=3, A039649(3)=3 - so there are three pairs in the fourth row.
Those pairs are supposed to have the same ratio, which can be calculated using the expression A221740(p)/A221741(p) = (p^2*(p+1)^p - (p+1)^p+1)/(-p^2 + p*(p+1)^p + (p+1)^p - p - 1), which for n=4 (p = n-1 = 3) yields 19/9 = 2.111.
The exhaustive computer program featured in the link finds that in decimal notation those three pairs with the ratio 19/9 = 2.111... are:
  (1) {114,54}; (2) {57,27}; (3) {228,108}
In base-4 notation, those 3 pairs of distinct permutations are:
  (1) {1302, 0312}; (2) {0321,0123}; (3) {3210,1230};
Now we calculate the fourth row terms per the sequence's definition:
  (1302-0123)/9 = 131; (0312-0123)/9 = 21; (0321-0123)/9 = 22; (0123-0123)/9 = 0; (3210-0123)/9 = 343; (1230-0123)/9 = 123;
Thus, for the fourth row (n=4), which corresponds to base 4 (note that rows in the table are counted starting with n=2, which corresponds to base 2) we get the following 6 (three pairs) sequence terms, presented as sorted in ascending order: 0, 21, 22, 123, 131, 343, ...

Links

Crossrefs

Cf. A215940, A221740, A221741.

A358314 Triangle T(n,k) read by rows where T(2m - 1,k) = (A051845(2m - 1,k))/(2m - 1) and T(2m,k) = A051845(2m,k)/m for m > 0, k > 0.

Original entry on oeis.org

1, 5, 7, 9, 10, 13, 15, 18, 19, 97, 99, 107, 111, 119, 121, 147, 149, 167, 173, 179, 183, 207, 211, 217, 223, 241, 243, 269, 271, 279, 283, 373, 374, 379, 381, 386, 387, 409, 410, 421, 424, 428, 430, 451, 453, 457, 460, 471
Offset: 1

Views

Author

Alexander R. Povolotsky, Nov 08 2022

Keywords

Comments

The n-th row has n! elements.

Examples

			Triangle begins:
       k=1  k=2  k=3 ...
  n=1:   1;
  n=2:   5,   7;
  n=3:   9,  10, ...,  19;
  n=4:  97,  99, 107, ..., 283;
  n=5: 373, 374, 379, 381, ..., 471;
  ...

Crossrefs

Cf. A051845, left edge = A221741, right edge = A221740.

Formula

T(n,1) = 4*(((n+1)^(n+1)-(n+1))/((n+1)-1)^2-1)/((3-(-1)^n)*n) = A221741(n).
T(n,n!) = 4*((n-1)*(n+1)^(n+1)+1)/((3-(-1)^n)*n^3) = A221740(n).
Showing 1-4 of 4 results.

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