sc.pp.recipe_zheng17 function is different from the Zheng et al. (2017) paper

The following is what this function does (we can see it with ?sc.pp.recipe_zheng17):

sc.pp.filter_genes(adata, min_counts=1)  # only consider genes with more than 1 count
sc.pp.normalize_per_cell(                # normalize with total UMI count per cell
         adata, key_n_counts='n_counts_all')
filter_result = sc.pp.filter_genes_dispersion(  # select highly-variable genes
        adata.X, flavor='cell_ranger', n_top_genes=n_top_genes, log=False)
adata = adata[:, filter_result.gene_subset]     # subset the genes
sc.pp.normalize_per_cell(adata)          # renormalize after filtering
if log: sc.pp.log1p(adata)               # log transform: adata.X = log(adata.X + 1)
sc.pp.scale(adata)                       # scale to unit variance and shift to zero mean

But in the original paper Zheng et al. (2017 (https://www.nature.com/articles/ncomms14049#Sec11), it said:
Only genes with at least one UMI count detected in at least one cell are used. UMI normalization was performed by first dividing UMI counts by the total UMI counts in each cell, followed by multiplication with the median of the total UMI counts across cells. Then, we took the natural log of the UMI counts. Finally, each gene was normalized such that the mean signal for each gene is 0, and standard deviation is 1.

So, comparing these two pipelines, the pipeline implemented in scanpy is not the same with the method described in the original paper, in the paper, there is a step: multiplication with the median of the total UMI counts across cells, but this step was skipped inside the function sc.pp.recipe_zheng17.

Is there anyone who can tell me why they are different? @flying-sheep

Start at sc.pp.recipe_zheng17 and read its documented pipeline, then compare each normalization and transformation step with Section 11 of the Zheng et al. (2017) paper. Check specifically whether multiplication by the median total UMI count is represented, and determine whether the outcome should explain the difference or align with the paper.